32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen. Q. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2 = n1 exp [-mg (h2 – h1)/ kBT] Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n2 = n1 exp [-mg NA(ρ - P′) (h2 –h1)/ (ρRT)] Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.] AnswerAccording to the law of atmospheres, we have: n2 = n1 exp [-mg (h2 – h1) / kBT] … (i) where, n1 is the number density at height h1, and n2 is the number density at height h2 mg is the weight of the particle suspended in the gas column Density of the medium = ρ' Density of the suspended particle = ρ Mass of one suspended particle = m' Mass of the medium displaced = m Volume of a suspended particle = V According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as: Weight of the medium displaced – Weight of the suspended particle = mg – m'g = mg - V ρ' g = mg - (m/ρ)ρ'g = mg(1 - (ρ'/ρ) ) ....(ii) Gas constant, R = kBN kB = R / N ....(iii) Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:
n2 = n1 exp [-mg (h2 – h1) / kBT]= n1 exp [-mg (1 - (ρ'/ρ) )(h2 – h1)(N/RT) ]= n1 exp [-mg (ρ - ρ')(h2 – h1)(N/RTρ) ]Q. Given below are densities of some solids and liquids. Give rough estimates of thesize of their atoms: [Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å]. Ans. If r is the radius of the atom, then volume of each atom = 4/3 π r3 Volume of all atoms in one mole of substance = 4/3 π r3 × N = M/ρ ∴ r = [ 3M / 4πρN]1/3For Carbon, -3M = 12.01 × 10 Kgρ = 2.22 × 3 Kg -3 10 mSimilarly,for gold, r = 1.59 Åfor liquid nitrogen, r = 1.77 Å for lithium, r = 1.73 Åfor liquid fluorine, r = 1.88 ÅQ. Which of the following examples represent periodic motion?(1) A swimmer completing one (return) trip from one bank of a river to the other and back.(2) A freely suspended bar magnet displaced from its N-S direction and released.(3) A hydrogen molecule rotating about its center of mass.(4) An arrow released from a bow.Ans.(a) The swimmer's motion is not periodic. Though the motion of a swimmer is to and fro but will not have adefinite period.(a) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic becausethe magnet oscillates about its position with a definite period of time.(m) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again andagain after an equal interval of time. Such a motion is periodic.(n) An arrow released from a bow moves only in the forward direction. It does not come backward.Hence, this motion is not a periodic.
14. Oscillations Q. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? A. the rotation of earth about its axis. B. motion of an oscillating mercury column in a U-tube. C. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. D.general vibrations of a polyatomic molecule about its equilibrium position. Answer A.It is priodic but not simple harmonic motion because it is not to and fro about a fixed point. B.It is a simple harmonic motion because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time. C.It is simple harmonic motion because the ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. (d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic. Q. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)Ans.
(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetitionof motion in this case(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a periodof 2 s.(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For aperiodic motion, the entire motion of the particle must be repeated in equal intervals of time.(d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having aperiod of 2 s.14.4. Which of the following functions of time represent (a) simple harmonic, (b) periodic but notsimple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω isany positive constant):A.sin ωt – cos ωt 3 ωtB.sin
C.3 cos (π/4 – 2ωt)D.cos ωt + cos 3ωt + cos 5ωtE.exp (–ω2t2)Answer(a) SHMThe given function is: sin ωt – cos ωtThis function represents SHM as it can be written in the form: a sin (ωt + Φ) Its period is: 2π/ω(b) Periodic but not SHM The given function is: 3 ωt = 1/4 [3 sin ωt - sin3ωt]sinThe terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, thesuperposition of two SHM is periodic and not simple harmonic.Ites period is: 2π/ω(c) SHMThe given function is:This function represents simple harmonic motion because it can be written in the form: a cos (ωt + Φ)Itsperiod is: 2π/2ω = π/ω(d) Periodic, but not SHMThe given function is cosωt + cos3ωt + cos5ωt. Each individual cosine function represents SHM.However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.(e) Non-periodic motionThe given function exp(-ω2t2) is an exponential function. Exponential functions do not repeat themselves.Therefore, it is a non-periodic motion.(f) The given function 1 + ωt + ω2t2 is non-periodic.14.5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take thedirection from A to B as the positive direction and give the signs of velocity, acceleration and force onthe particle when it is
A.at the end A,B.at the end B,C.at the mid-point of AB going towards A,D.at 2 cm away from B going towards A,E.at 3 cm away from A going towards B, andF.at 4 cm away from B going towards A.Ans.From above figure, where A and B represent the twoextreme positions of a SHM. FOr velocity, the direction from A to B is taken to b positive. The acceleration andthe force, along AP are taken as positive and alsong Bp are taken as negative.(a) At the end A, the particle executing SHM is momentarily at rest being its extreme position of motion. Therefore,its velocity is zero. Acceleration is positive becaue it is directed along AP, Force is also Positive since the force isdirected along AP.B.At the end B, velocity is zero. Here, accleration and force are negative as they are directed along BP.C.At the mid point of AB going towards A, the particle is at its mean psoiton P, with a tendancy to move along PA.Hence, velocity is positive. Both accleration and force are zero.D.At 2 cm away from B going towards A, the particle is at Q, with a tendancy to move along QP, which isnegative direction. Therefore, velocity, acceleration and force all are positive.E.At 3 sm away from A going towards B, the particle is at R, with a tendancy to move along RP, which ispositive direction. Here, velocity, acceleration all are positive.F.At 4 cm away from A going towards A, the particles is at S, with a tendancy to move along SA, which isnegative direction. Therefore, velocity is negative but acclereation is directed towards mean position,along SP. Hence it is positive and also force is positive similarly.Q.. Which of the following relationships between the acceleration a and the displacement x ofa particle involve simple harmonic motion?A.a = 0.7xB.a = –200x2C.a = –10x 3D.a = 100xAns.ted automaticalIn SHM, acceleration a is related to displacement by the relation of the form a = -kx, which is forrelation (c)
Q. The motion of a particle executing simple harmonic motion is described by thedisplacement function,x (t) = A cos (ωt + φ).If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s,what are its amplitude and initial phase angle? The angular frequency of theparticle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phaseof the particle with the above initial conditions.Ans.Intially, at t = 0; Displacement, x = 1 cm Intial velocity, v = ω cm/ sec.Angular frequency, ω = π rad/s–1 It is given that,x(t) = A cos(ωt + Φ)1 = A cos(ω × 0 + Φ) = A cos ΦA cosΦ = 1 ...(i)Velocity, v= dx/dtω = -A ωsin(ωt + Φ)1 = -A sin(ω × 0 + Φ) = -A sin Φ A sin Φ = -1 ...(ii)Squaring and adding equations (i) and (ii), we get: 2 2 Φ + 2 Φ) = 1 + 12 A (sin cosA =2∴ A = √2 cmDividing equation (ii) by equation (i), we gettanΦ = -1∴ Φ = 3π/4 , 7π/4,......SHM is given as: x = Bsin (ωt + α)Putting the given values in this equation, we get: 1 = Bsin[ω × 0 + α] = 1 + 1Bsin α = 1 ...(iii)Velocity, v = ωBcos (ωt + α) Substituting the given values, we get:π = πBsin αBsin α = 1 ...(iv)Squaring and adding equations (iii) and (iv), we get: 2 2 α + 2 α] = 1 + 12 B [sin cosB =2∴ B = √2 cmDividing equation (iii) by equation (iv), we get: Bsin α / Bcos α = 1/1tan α = 1 = tan π/4∴ α = π/4, 5π/4,......Q. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. Abody suspended from this balance, when displaced and released, oscillates with a period of 0.6 s.What is the weight of the body?
AnswerMaximum mass that the scale can read, M = 50 kgMaximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 mTime period, T = 0.6 sMaximum force exerted on the spring, F = Mg where, 2g = acceleration due to gravity = 9.8 m/s F = 50 × 9.8 = 490 -1∴ Spring constant, k = F/l = 490/0.2 = 2450 N m. Mass m, is suspended from the balance.∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N Hence, the weight of the body is about 219 N. –114.9. A spring having with a spring constant 1200 N m is mounted on a horizontal table as shown inFig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to adistance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) themaximum speed of the mass.Answer –1Spring constant, k = 1200 N mMass, m = 3 kg
PDDisplacement, A = 2.0 cm = 0.02 cm(i) Frequency of oscillation v, is given by the relation:Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation: a = ω2 Awhere, 2Hence, the maximum acceleration of the mass is 8.0 m/s .(iii) Maximum velocity, vmax = AωHence, the maximum velocity of the mass is 0.4 m/s.Q.In Exercise 14.9, let us take the position of mass when the spring is unstrechedas x = 0, and the direction from left to right as the positive direction of x-axis. Givex as a function of time tfor the oscillating mass if at the moment we start the stopwatch (t = 0), the mass isA.at the mean position,B.at the maximum stretched position, andC.at the maximum compressed position.In what way do these functions for SHM differ from each other, in frequency, inamplitude or the initial phase?AnswerDistance travelled by the mass sideways, a = 2.0 cm Angular frequency of oscillation:
(a) As time is noted from the mean position, hence using x = a sin ω t, we have x = 2 sin 20 t(b) At maximum stretched position, the body is at the extreme right position, with an intial phase of π/2 rad.Then,(c) At maximum compressed position, the body is at left position, with an intial phase of 3 π/2 rad.Then,The functions neither differ in amplitude nor in frequency. They differ in intial phase.14.11. Figures 14.29 correspond to two circular motions. The radius of the circle, the period ofrevolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) areindicated on each figure.Obtain the corresponding simple harmonic motions of the x-projection of the radius vector ofthe revolving particle P, in each case.Ans.(a) Time period, t = 2 s Amplitude, A = 3 cmAt time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis,.e., phase angle Φ = +π/2Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given bythe displacement equation:
(b) Time Period, t = 4 s Amplitude, a = 2 mAt time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction, Hence, phase angleΦ = +π Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, isgiven as:14.12. Plot the corresponding reference circle for each of the following simple harmonic motions.Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed ofthe rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in everycase: (x is in cm and t is in s).= x = –2 sin (3t + π/3)= x = cos (π/6 – t)= x = 3 sin (2πt + π/4)= x = 2 cos πtAns.(a)Amplitude, A = 2 cmPhase angle, Φ = 5π/6 = 150°.Angular velocity = ω = 2π/T = 3rad/sec.The motion of the particle can be plotted as shown in fig. 10(a).(b)
Amplitude, A = 1Phase angle, Φ = -π/6 = -30°. Angular velocity, ω = 2π/T = 1 rad/s.The motion of the particle can be plotted as shown in fig. 10(b).(c)Amplitude, A = 3 cmPhase angle, Φ = 3π/4 = 135°Angular velocity, ω = 2π/T = 2 rad/s.(d)The motion of the particle can be plotted as shown in fig. 10(c).Amplitude, A = 2 cm Phase angle, Φ = 0Angular velocity, ω = π rad/s.The motion of the particle can be plotted as shown in fig. 10(d).14.13. Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass mattached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) showsthe same spring with both ends free and attached to a mass m at either end. Each end of the spring inFig. 14.30(b) is stretched by the same force F.(a) What is the maximum extension of the spring in the two cases?(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation ineach case?
Answer(a) The maximum extension of the spring in both cases will = Flk, where k is the spring constant of thesprings used.(b) In Fig.14.30(a) if x is the extension in the spring, when massm is returning to its mean position after being released free, then restoring force on the mass iss F = -kx, i.e., F ∝ xAs, this F is directed towards mean position of the mass, hence the mass attached to the spring willexecute SHM.Spring factor = spring constant = kinertia factor = mass of the given mass = m As time period,In Fig.14.30(b), we have a two body system of spring constant k and reduced mass, µ = m × m / m + m =m/2.Inertia factor = m/2 Spring factor = k14.14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m.If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, whatis its maximum speed?Ans.Angular frequency of the piston, ω = 200 rad/ min. Stroke = 1.0 mAmplitude, A = 1.0/2 = 0.5 mThe maximum speed (vmax) of piston is given by the relation: vmax = Aω = 200 × 0.5 = 100m/min.14.15. The acceleration due to gravity on the surface of moon is 1.7 ms–2. Whatis the time period of a simple pendulum on the surface of moon if its time periodon the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2) –Ans.Acceleration due to gravity on the surface of moon,g' = 1.7 m s 2 –2Acceleration due to gravity on the surface of earth, g = 9.8 m s Time period of a simple pendulum on earth,T = 3.5 s
Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.14.16. Answer the following questions:(a) Time period of a particle in SHM depends on the force constant k and mass m of theparticle:T = 2π √m/k. A simple pendulum executes SHM approximately. Why then is the time periodof a pendulum independent of the mass of the pendulum?(b) The motion of a simple pendulum is approximately simple harmonic for small angleoscillations. For larger angles of oscillation, a more involved analysis shows that T isgreater than 2π √l/gThink of a qualitative argument to appreciate this result.(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch givecorrect time during the free fall?(d) What is the frequency of oscillation of a simple pendulum mounted in a cabinthat isfreely falling under gravity?Ans.(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, mcancels out in denominator as well as in numerator. That is why the time period of simple pendulum isindependent of the mass of the bob.(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:F = –mg sinθ where,F = Restoring force m = Mass of the bobg = Acceleration due to gravity θ = Angle of displacementFor small θ, sinθ ≈ θ
For large θ, sinθ is greater than θ. This decreases the effective value of g. Hence, the timeperiod increases as:T = 2π √l/g(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do withgravity.(d) Gravity disappears for a man under free fall, so frequency is zero.14.17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car ismoving on a circular track of radius R with a uniform speed v. If the pendulum makes smalloscillations in a radial direction about its equilibrium position, what will be its time period?Ans.The bob of the simple pendulum will experience the acceleration due to gravity and the centripetalacceleration provided by the circular motion of the car.Acceleration due to gravity = g 2Centripetal acceleration = v /R where,v is the uniform speed of the car R is the radius of the trackEffective acceleration (g') is given as:Q.Cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1. Thecork is depressed slightly and then released. Show that the cork oscillates up and down simpleharmonically with a period where ρ is the density of cork. (Ignore damping due to viscosity of theliquid).Ans.Base area of the cork = AHeight of the cork = hDensity of the liquid = ρ1Density of the cork = ρIn equilibrium:Weight of the cork = Weight of the liquid displaced by the floating corkLet the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced.Hence, an extra up-thrust acts upward and provides the restoring force to the cork. Up-thrust = Restoringforce, F = Weight of the extra water displacedF = –(Volume × Density × g)Volume = Area × Distance through which the cork is depressed Volume = Ax∴ F = – A x ρ1 g .....(i) Accroding to the force law:F = kx k = F/xwhere, k is constant k = F/x = -Aρ1 g ....(ii)
The time period of the oscillations of the cork:T = 2π √m/k ....(iii) where,m = Mass of the cork=Volume of the cork × Density=Base area of the cork × Height of the cork × Density of the cork=AhρHence, the expression for the time period becomes:--Q. One end of a U-tube containing mercury is connected to a suction pump and the other endto atmosphere. A small pressure difference is maintained between the two columns. Showthat, when the suction pump is removed, the column of mercury in the U-tube executessimple harmonic motion.Ans.Area of cross-section of the U-tube = A Density of the mercury column = ρAcceleration due to gravity = gRestoring force, F = Weight of the mercury column of a certain height= = –(Volume × Density × g)== –(A × 2h × ρ × g) = –2Aρgh = –k × Displacement in one of the arms (h)Where,2h is the height of the mercury column in the two arms k is a constant, given by k = -F/h = 2Aρgwhere,M is mass of mercury columnLet l be the length of the total mercury in the U-tube.Mass of mercury, m = Volume of mercury × Density of mercury =AlρHence, the mercury column executes simple harmonic motion with time period 2π √l/2g .Q.An air chamber of volume V has a neck area of cross section a into which a ball of mass m justfits and can move up and down without any friction (Fig.14.33). Show that when the ball ispressed down a little and released, it executes SHM. Obtain an expression for the time period ofoscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.33].
Ans.Volume of the air chamber = VArea of cross-section of the neck = a Mass of the ball = mThe pressure inside the chamber is equal to the atmospheric pressure.Let the ball be depressed by x units. As a result of this depression, there would be a decrease in thevolume and an increase in the pressure inside the chamber.Decrease in the volume of the air chamber, V = ax Volumetric strain = Change in volume/original volume⇒ V/V = ax/VBulk modulus of air, B = Stress/Strain = -p/ax/VIn this case, stress is the increase in pressure. The negative sign indicates that pressure increases with decrease involume.p = -Bax/VThe restoring force acting on the ball, F = p × a= -Bax/V .a 2= -Bax /V ......(i)In simple harmonic motion, the equation for restoring force is: F = -kx .....(ii) 2where, k is the spring constant Comparing equations (i) and (ii), we get: k = Ba /VTime Period,Q.You are riding in an automobile of mass 3000 kg.Assuming that you areexamining the oscillation characteristics of its suspension system. Thesuspension sags 15 cm when the entire automobile is placed on it. Also, the
amplitude of oscillation decreases by 50% during one complete oscillation.Estimate the values of (a) the spring constant k and(b) the damping constant b for the spring and shock absorber system of onewheel, assuming that each wheel supports 750 kg.Ans.(a) Mass of the automobile, m = 3000 kgDisplacement in the suspension system, x = 15 cm = 0.15 m There are 4 springs in parallel to the supportof the mass of the automobile.The equation for the restoring force for the system:F = –4kx = mgWhere, k is the spring constant of the suspension system Time period, T = 2π √m/4k 44and k = mg/4x = 3000 × 10/ 4 × 0.15 = 5000 = 5 × 10 Nm Spring Constant, k = 5 × 10 Nm(b) Each wheel supports a mass, M = 3000/4 = 750 kgFor damping factor b, the equation for displacement is written as: -bt/2Mx = x0eThe amplitude of oscilliation decreases by 50 %. ∴ x = x0/2 -bt/2Mx0/2 = x0eloge2 = bt/2M∴ b = 2M loge2 / twhere,Q. Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equalsthe average potential energy over the same period.AnswerThe equation of displacement of a particle executing SHM at an instant t is given as:x = Asin ωt where,A = Amplitude of oscillation ω = Angular frequency = √k/MThe velocity of the particle is:v = dx/dt = Aωcosωt The kinetic energy of the particle is:Ek = 1/2 2 = 1/2 2 ω2 2 ωt Mv MA cosThe portential energy of the particle is: Ep = 1/2 2 = 1/2 2 ω2 2 2 ωt kx M A sinFor time period T, the average kinetic energy over a single cycle is given asAnd, average potential energy over one cycle is given as:
Average potential energy over one cycle is given as
It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal tothe average potential energy for the same time period.14.23. A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twistedby rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radiusof the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant αis defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).Ans.Mass of the circular disc, m = 10 kg Radius of the disc, r = 15 cm = 0.15 mThe torsional oscillations of the disc has a time period, T = 1.5 s The moment of inertia of the disc is: 2I = 1/2mr 2= 1/2 × (10) × (0.15)= 0.1125 kg/m2Time Period, T = 2π √I/α= is the torisonal constant.= = 4π2 I / 2 2 (π)2 T (1.5)= 4 × × 0.1125 /= 1.972 Nm/rad –1Hence, the torsional spring constant of the wire is 1.972 Nm rad .Q. A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Findthe acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.Answerr = 5 cm = 0.05 m T = 0.2 sω = 2π/T = 2π/0.2 = 10 π rad/sWhen displacement is y, then acceleration, A = -ω2 yCase (a) When y = 5cm = 0.05 mA = -(10π)2 × 0.05 = -5π2 2 m/sCase (b) When y = 3 cm = 0.03 mA = -(10π)2 × 0.03 = -3π2 2 m/sCase (c) When y = 0A = -(10π)2 × 0 = 0
Q. A mass attached to a spring is free to oscillate, with angular velocity ω, in ahorizontal plane without friction or damping. It is pulled to a distance x0 andpushed towards thecentre with a velocity v0 at time t = 0. Determine the amplitude of the resultingoscillations in terms of the parameters ω, x0 and v0. [Hint: Start with theequation x = a cos (ωt+θ) and note that the initial velocity is negative.]AnswerThe displacement rquation for an oscillating mass is given by:== Acos(ωt + θ) where,A is the amplitude=is the displacement θ is the phase constantVelcoity, v = dx/dt = -Aωsin(ωt + θ) At t = 0, x = x0x0 = Acos θ = x0 ....(i) and, dx/dt = -v0 = Aωsinθ Asinθ = v0/ω ...(ii)LSquaring and adding equations (i) and (ii), we get:PDF generated automatically by the HTML to PDF API of PDFmyURL
15.WAVESQ. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbancetake to reach the other end?Ans.Mass of the string, M = 2.50 kg Tension in the string, T = 200 N Length of the string, l =20.0 m -1Mass per unit length, µ = M/l = 2.50/20 = 0.125 kg mThe velocity (v) of the transverse wave in the string is given by the relation:v = √t/µ= √200/0.125 = √1600 = 40 m/s∴ Time taken by the disturbance to reach the other end, t = l/v = 20/40 = 0.5 s.Q. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pondnear the base of the tower. When is the splash heard at the top given that the speed of sound in air is –1 –2340 m s ? (g= 9.8 m s )AnswerHeight of the tower, s = 300 m Initial velocity of the stone, u = 0 2Acceleration, a = g = 9.8 m/s Speed of sound in air = 340 m/sThe time (t1) taken by the stone to strike the water in the pond can be calculated using the second equation ofmotion, as: 2s = ut1 + 1/2 gt1 2300 = 0 + 1/2 × 9.8 × t1∴ t1 = √300 × 2/9.8 = 7.82 sTime taken by the sound to reach the top of the tower, t2 =300/340 = 0.88 sTherefore, the time after which splash is heard, t = t1 + t2 = 7.82 + 0.88 = 8.7 s.Q. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in thewire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C –1= 343 m s .AnswerLength of the steel wire, l = 12 m Mass of the steel wire, m = 2.10 kgVelocity of the transverse wave, v = 343 m/s -1Mass per unit length, µ = m/l = 2.10/12 = 0.175 kg mFor Tension T, velocity of the transverse wave can be obtained using the relation:v = √T/µ ∴ T = vu= 2 × 0.175 = 20588.575 ≈ 2.06 × 104 N. (343)
15.4. Use the formula v = √γP/ρ to explain why the speed of sound in air(i) is independent of pressure,(j) increases with temperature,(k) increases with humidity.Ans.(a) Take the relation:v = √ γP/ρ ....(i)where,Density, ρ = Mass/Volume = M/VM = Molecular weight of the gas V = Volume of the gasHence, equation (i) reduces to: v = √γPV/M ....(ii)Now from the ideal gas equation for n = 1:PV = RTFor constant T, PV = ConstantSince both M and γ are constants, v = ConstantHence, at a constant temperature, the speed of sound in a gaseous medium is independent of thechange in the pressure of the gas.(b) Take the relation: v = √γP/ρ ....(i)For one mole of any ideal gas, the equation can be written as:PV = RTP = RT/V ....(ii)Substituting equation (ii) in equation (i), we get: v = √γRT/Vρ = √γRT/M .....(iii)where,mass, M = ρV is a constant γ and R are also constantsWe conclude from equation (iii) that v ∝ √THence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseousmedium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous mediumand vice versa.(c) Let vm and vd be the speed of sound in moist air and dry airRespectively Let ρm and ρd be the densities of the moist air and dry air respectively
Hρoweρver, the presence of water vapour reduces the density of air, i.e., d< m∴ vm > vdHence, the speed of sound in mois air is greater than it is in dry air. Thus, in gaseous medium, the speedof sound increases with humidity.Q. You have learnt that a travelling wave in one dimension is represented by afunction y = f (x, t)where x and t must appear in the combination x – v t or x + v t, i.e.y = f (x ± v t). Is the converse true? Examine if the following functions for y canpossibly represent a travelling wave:(o) (x – vt)2(p) log [(x + vt) / x0](q) 1 / (x + vt)AnswerNo, the converse is not true. The basic requirements for a wave function to represent a travelling wave isthat for all values of x and t, wave function must have finite value.Out of the given functions for y, no one satisfies this condition. Therefore, none can represent a travellingwave.Q. A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets awater surface, what is the wavelength of(a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 ms–1 and in water 1486 m s–1.Answer(a) Frequency of the ultrasonic sound, ν = 1000 kHz = 6 Hz Speed of sound in air, va = 340 m/s 10The wavelength (λr) of the reflected sound is given by therelation: λr = v/v 6 -4= 340/10 = 3.4 × 10 m.(b) Frequency of the ultrasonic sound, ν = 1000 kHz = 6 Hz Speed of sound in water, vw = 1486 m/s 10The wavelength of the transmitted sound is given as: λr = 1486 × 6 = 1.49 × -3 m. 10 1015.7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. Whatis the wavelength of sound in the tissue in whichthe speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2MHz.Ans. 3Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 10 m/s
Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz The wavelength of sound in the tissue isgiven as:λ = v/v 36 -4= 1.7 × 10 / 4.2 × 10 = 4.1 × 10 m.15.8. A transverse harmonic wave on a string is described byWhere x and y are in cm and t in s. The positive direction of x is from left to right.(a) Is this a travelling wave or a stationary wave?If it is travelling, what are the speed and direction of its propagation?(b) What are its amplitude and frequency?(c) What is the initial phase at the origin?(d) What is the least distance between two successive crests in the wave?Ans.(a) The equation of progressive wave travelling from right to left is given by the displacementfunction:y (x, t) = a sin (ωt + kx + Φ) ....(i) The given equation is:On compaaring both the equations, we find that equation (ii) represents a travelling wave, propgatingfrom right to left. Now using equations (i) and (ii), we can write:ω = 36 rad/s and k = 0.018 m-1 We know that:v = ω/2π and λ = 2π/k Also,v = vλ∴ v = (ω/2π) / (2π/k) = ω/k= 36/0.018 = 2000 cm/s = 20 m/sHence, the speed of the given travelling wave is 20 m/s.(b) Amplitude of the given wave, a = 3 cm Frequency of the given wave:v = ω/2π = 36 / 2 × 3.14 = 573 Hz(c) On comparing equations (i) and (ii), we find that the intial phase angle, Φ = π/4(d) The distance between two successive crests or troughs is equal to the wavelength of the wave.Wavelength is given by the relation: k = 2π/λ∴ λ = 2π/k = 2 × 3.14 / 0.018 = 348.89 cm =3.49 m.Q. For the wave described in last question, plot the displacement (y) versus (t) graphs for x =0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatorymotion in travelling wave differ from one point to another: amplitude, frequency or phase?Ans.All the waves have different phases.The given transverse harmonic wave is:For x = 0, the equation reduces to:
Also,ω = 2π/t = -1 ∴ t = π/18 s 36rad/sNow, plotting y vs. t graphs using the different values of t, as listed in the given table t (s) 0 T/8 2T/7 3T/8 4T/8 5T/8 6T/8 7T/8 y 3 3/√2 -- 0 3/√2 0 3/√2 –3 3/√2 (cm)For x = 0, x = 2, and x = 4, the phases of the three waves will get changed. This is because amplitudeand frequency are invariant for any change in x. The y-t plots of the three waves are shown in the givenfigure.Q. For the travelling harmonic wave y (x, t) = 2.0 cos 2π(10t – 0.0080x + 0.35)Where x and y are in cm and t in s. Calculate the phase differencebetween oscillatory motion of two points separated by a distance ======== 4m= 0.5 m,= λ/2= 3λ/4AnswerEquation for a travelling harmonic wave is given as: y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)= 2.0 cos (20πt – 0.016πx + 0.70 π) Where,
Propagation constant, k = 0.0160 π Amplitude, a = 2 cmAngular frequency, ω= 20 π rad/s Phase difference is given by the relation: Φ = kx =2π/λ(a) For x = 4 m = 400 cmΦ = 0.016 π × 400 = 6.4 π rad(b) For 0.5 m = 50 cmΦ = 0.016 π × 50 = 0.8 π rad(c) For x = λ/2Φ = 2π/λ × λ/2 = π rad(d) For x = 3λ/4Φ = 2π/λ × 3λ/4 = 1.5π rad.15.11. The transverse displacement of a string (clamped at its both ends) isgiven byWhere x and y are in m and t in s. The length of the string is 1.5m and its mass is 3.0 ×10–2 kg. Answer the following:= Does the function represent a travelling wave or a stationary wave?= Interpret the wave as a superposition of two waves travelling in oppositedirections. What is the wavelength, frequency, and speed of each wave?= Determine the tension in the string.Ans.(a) The general equation representing a stationary wave is given by the displacement function:y (x, t) = 2a sin kx cos ωtThis equation is similar to the given equation:Hence, the given equation represents a stationary wave.(b) A wave travelling along the positive x-direction is given as: y1 = a sin (ωt - kx)The wave travelling along the positive x-direction is given as: y2 = a sin (ωt + kx)The supposition of these two waves yields: y = y1 + y2 = a sin (ωt - kx) - a sin (ωt + kx)=a sin (ωt) cos (kx) - a sin (kx) cos (ωt) - a sin (ωt) cos (kx) - a sin (kx) cos (ωt)= 2 a sin (kx) cos (ωt)
∴Wavelength, λ = 3 m It is given that:120π = 2πν Frequency, ν = 60 Hz Wave speed, v = νλ = 60 × 3 = 180 m/s(c) The velocity of a transverse wave travelling in a string is given by the relation:v = √T/µ ....(i) where,Velocity of the transverse wave, v = 180 m/s –2Mass of the string, m = 3.0 × 10 kg Length of the string, l = 1.5 mMass per unit length of the string, µ = m/l -2= 3.0 × 1.5 = 10 -2 -1= 2 × 10 kg m Tension in the string = TFrom equation (i), tension can be obtained as:T = v2μ –2 2= (180) × 2 × 10= 648 NQ. (i) For the wave on a string described in Exercise 15.11, do all the points on the stringoscillate with the same (a)frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of apoint 0.375 m away from one end?Ans.All the points on the string= have the same frequency except at the nodes (where frequency is zero)= have thse same phase everywhere in one loop except at the nodes.However, the amplitude of vibration at different points is different.Q. Given below are some functions of x and t to represent the displacement (transverse orlongitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) astationary wave or (iii) none at all:(a) y = 2 cos (3x) sin (10t)(b) y = 2 √x - vt
(c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)(d) y = cos x sin t + cos 2x sin 2tAnswer(a) The given equation represents a stationary wave because the harmonic terms kx and ωt appearseparately in the equation.(b) The given equation does not contain any harmonic term. Therefore, it does not represent either atravelling wave or a stationary wave.(c) The given equation represents a travelling wave as the harmonic terms kx and ωt are in the combination of kx– ωt.(d) The given equation represents a stationary wave because the harmonic terms kx and ωt appearseparately in the equation. This equation actually represents the superposition of two stationary waves.Q.. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of –2 –2the wire is 3.5 × 10 kg and its linear mass density is 4.0 × 10 –1kg m . What is (a) the speed of a transverse wave on the string, and (b) the tension in thestring?Ans. –2(a) Mass of the wire, m = 3.5 × 10 kg -2 -1Linear mass density, μ = m/l = 4.0 × 10 kg m wire, l = m/μ = 3.5 × –2 / 4.0 × -2 = 0.875 mFrequency of vibration, v = 45 Hz∴ length of the 10 10The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:= = 2l/m where,n = Number of nodes in the wire For fundamental node, n = 1:= = 2l= = 2 × 0.875 = 1.75 mThe speed of the transverse wave in the string is given as:v = νλ= 45 × 1.75 = 78.75 m/s(b) The tension produced in the string is given by the relation: 2T=v µ 2 –2= (78.75) × 4.0 × 10 = 248.06 NQ. A metre-long tube open at one end, with a movable piston at the other end, shows resonancewith a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cmor 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edgeeffects may be neglected.AnswerFrequency of the turning fork, ν = 340 HzSince the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed andthe other end open, as shown in the given figure.
Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation:l1 = π/4 where,length of pipe, l1 = 25.5 cm = 0.255 m∴ λ = 4l1 = 4 × 0.255 = 1.02 mThe speed of the sound is given by the relation: v = vλ = 340 × 1.02 = 346.8m/s.Q. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrationsof the rod is given to be 2.53 kHz. What is the speed of sound in steel?AnswerLength of the steel rod, l = 100 cm = 1 mFundamental frequency of vibration, ν = 2.53 kHz = 2.53 × 103 HzWhen the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed atits two ends, as shown in the given figure.The distance between two successive node is λ/2 ∴ l = λ/2λ = 2l = 2 × l = 2 mThe speed of sound in steel is given by the relation: v = νλ 3= 2.53 × 10 × 2 3= 5.06 × 10 m/s= 5.06 km/s15.17. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430Hz source? Will the same source be in resonance with the pipe if both ends are –1open? (Speed of sound in air is 340 m s ).Answer
First (Fundamental); NoLength of the pipe, l = 20 cm = 0.2 mSource frequency = th normal mode of frequency, νn = 430 Hz Speed of sound, v = 340 m/s n thIn a closed pipe, the n normal mode of frequency is given by the relation:Hence, the first mode of vibration frequency is resonantly excited by the given source. thIn a pipe open at both ends, the n mode of vibration frequency is given by the relation:vn = nv/2l n = 2lvn/v15.18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats offrequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found toreduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?AnswerFrequency of string A, fA = 324 HzFrequency of string B = fBBeat’s frequency, n = 6 HzBeat's Frequency is given as:Frequency decreases with a decrease in the tension in a string. This is because frequency is directlyproportional to the square root of tension. It is given as:v ∝ √THence, the beat frequency cannot be 330 Hz ∴ fB = 318 Hz.15.19. Explain why (or how):= In a sound wave, a displacement node is a pressure antinode and vice versa,= Bats can ascertain distances, directions, nature, and sizes of the obstacles without any“eyes”,= A violin note and sitar note may have the same frequency, yet we can distinguish between thetwo notes,= Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagatein gases, and= The shape of a pulse gets distorted during propagation in a dispersive medium.
Answer= A node (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum.An antinode (A) is a point where the amplitude of vibration is the maximum and pressure is theminimum.Therefore, a displacement node is nothing but a pressure antinode, and vice versa.= Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in theirpath, they givethem the idea about the distance, direction, size and nature of the obstacle.= The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence,one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency ofvibration.= This is because solids have both, the elasticity of volume and elasticity of shape, whereas gaseshave only the volume elasticity.= A sound pulse is a combination of waves of different wavelength. As waves of different λ travel in adispersive medium with different velocities, therefore, the shape of the pulse gets distorted.15.20. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in stillair. (i) What is the frequency of the whistle for a platform observer when the train –1(a) approaches the platform with a speed of 10 m s , (b) –1recedes from the platform with a speed of 10 m s ? (ii) What is the speed of sound in each case? –1The speed of sound in still air can be taken as 340 m s .Answer(i) (a)Frequency of the whistle, ν = 400 Hz Speed of the train, vT= 10 m/sSpeed of sound, v = 340 m/sThe apparent frequency (v') of the whistle as the train approaches the platform is given by therelation:
(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and theobserver. These relative motions produce no effect on the speed of sound. Therefore, the speed of soundin air in both the cases remains the same, i.e., 340 m/s.15.21. A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind startsblowing in the –direction from the yard to the station with at a speed of 10 m s1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’splatform? Is the situation exactly identical to the case when the air is still and the –1observer runs towards the yard at a speed of 10 m s ? The speed of sound in still air can be taken as –1340 m s .Ans.For the stationary observer:Frequency of the sound produced by the whistle, ν = 400 Hz Speed of sound = 340 m/sVelocity of the wind, v = 10 m/sAs there is no relative motion between the source and the observer, the frequency of the sound heard by theobserver willbe the same as that produced by the source, i.e., 400 Hz.The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units,i.e.,Effective speed of the sound, ve = 340 + 10 = 350 m/sThe wavelength (λ) of the sound heard by the observer is given by the relation:λ = ve/v = 350/400 = 0.875 mFor the running observer:Velocity of the observer, vo = 10 m/sThe observer is moving toward the source. As a result of the relative motions of the source and theobserver, there is a change in frequency (v').This is given by the relation:Since the air is still, the effective speed of sound = 340 + 0 = 340 m/sThe source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m.Hence, the given two situations are not exactly identical.15.22. A travelling harmonic wave on a string is described by=What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is thisvelocity equal to the velocity of wave propagation?
= Locate the points of the string which have the same transverse displacements and velocity as the x = 1cm point at t= 2 s, 5 s and 11 s.Answer(a) The given harmonic wave is= 90 coss (732.81°) = 90 cos (90 × 8 + 12.81°)= 90 cos (12.81°)= 90 × 0.975 =87.75 cm/sNow, the equation of a propagating wave is given by: y (x, t) = asin (kx + wt + Φ)where, k = 2π/λ∴ λ = 2π/k and, ω = 2πv ∴ v = ω/2πSpeed, v = vλ = ω/k where,ω = 12rad/s k = 0.0050 -1 m∴ v = 12/0.0050 = 2400 cm/sHence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wavepropagation.(b) Propagation constant is related to wavelength as: k = 2π/λ∴ λ = 2π/k = 2 × 3.14 / 0.0050 = 1256 cm = 12.56 m
Therefore, all the points at distance nλ (n = ±1, ±2, .... and so on), i.e. ± 12.56 m, ± 25.12 m, … and so onfor x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.15.23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a)Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If thepulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s),is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?Answer(a) A short pip be a whistle has neither a definite wavelength nor a definite frequency. However, its speed ofpropagation is fized, being equal to speed of sound in air.(b) No, frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is thefrequency of repetition of the short pip of the whistle. –315.24. One end of a long string of linear mass density 8.0 × 10 –1kg m is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passesover a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incomingenergy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) ofthe string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. Theamplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t thatdescribes the wave on the string.AnswerThe equation of a travelling wave propagating along the positive y-direction is given by thedisplacement equation: y (x, t) = a sin (wt – kx) … (i)Linear mass density, μ = 8.0 × 10-3 kg m-1 Frequency of the tuning fork, ν = 256 HzAmplitude of the wave, a = 5.0 cm = 0.05 m … (ii) Mass of the pan, m = 90 kgTension in the string, T = mg = 90 × 9.8 = 882 NThe velocity of the transverse wave v, is given by the relation:Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacementequation:
y (x, t) = 0.05 sin (1.6 × 3 – 4.84 x) m. 10 t15.25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine movestowards the –1SONAR with a speed of 360 km h . What is the frequency of sound reflected by the submarine? –1Take the speed of sound in water to be 1450 m s .AnswerOperating frequency of the SONAR system, ν = 40 kHz Speed of the enemy submarine, ve = 360 km/h = 100 m/sSpeed of sound in water, v = 1450 m/sThe source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency (v')received and reflected by the submarine is given by the relation:Q. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both –1transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s , –1and that of P wave is 8.0 km s . A seismograph records P and S waves from an earthquake. The firstP wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at whatdistance does the earthquake occur?AnswerLet vSand vP be the velocities of S and P waves respectively.Let L be the distance between the epicentre and the seismograph.We have:L = vStS ...(i)L = vPtP ...(ii)Where,
tS and tP are the respective times taken by the S and P waves to reach the seismograph from theepicentreIt is given that: vP = 8 km/svS = 4 km/sFrom equations (i) and (ii), we have: vS tS = vP tP4tS = 8 tPtS = 2 tP ...(iii)It is also given that: tS – tP = 4 min = 240 s 2tP – tP = 240tP = 240And tS = 2 × 240 = 480 s From equation (ii), we get: L = 8 × 240= 1920 kmHence, the earthquake occurs at a distance of 1920 km from the seismograph.15.27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emissionfrequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is movingat 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?AnswerUltrasonic beep frequency emitted by the bat, ν = 40 kHz Velocity of the bat, vb = 0.03 vwhere, v = velocity of sound in airThe apparent frequency of the sound striking the wall is given as:
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