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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A man generates a symmetrical pulse in a string by moving his hand up and down. At `t=0` the point in his hand moves downward. The pulse travels with speed of `3 m//s` on the string & his hands passes `6` times in each second from the mean position. Then the point on the string at a distance `3m` will reach its upper extreme first time at time `t=` |
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Answer» Correct Answer - 15 `2f = 6, f = 3`, `T = (1)/(3)` sec, `v = 3 m//sec` `3 = (lambda)/(T) rArr lambda = 1m` `x = vt` `3 = 3t rArr t = 1` total time `= t + (3T)/(4) = 1 + (3)/(4) ((1)/(3)) = (5)/(4) = 1.25` sec. |
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| 2. |
The equation `y=A cos^(2) (2pi nt -2 pi (x)/(lambda))` represents a wave withA. Amplitude `A//2` , frequency 2n and wavelength `lambda//2`B. Amplitude `A//2` , frequency 2n and wavelength `lambda`C. Amplitude A , frequency 2n and wavelength `2lambda`D. Amplitude A , frequency n and waves |
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Answer» Correct Answer - a |
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| 3. |
A long string having a cross- sectional area `0.80 mm^2` and density `12.5 g cm^(-3)` is subjected to a tension of 64 N along the X-axis. One end of this string is attached to a vibrator moving in transverse direction at a frequency fo 20Hz. At t = 0. the source is at a maximujm displacement y= 1.0 cm. (a) Write the equation for the wave. (c ) What is the displacement of the particle of the string at x = 50 cm at time t = 0.05 s ? (d) What is the velocity of this particle at this instant ?A. `10sqrt(2)pi cm//s`B. `40sqrt(2) pi cm//s`C. `30sqrt(2)pi cm//s`D. `20sqrt(2) pi cm//s` |
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Answer» Mass per unit length of the string is `mu=Ad=(0.80 mm^(2))xx(12.5 g//cm^(3))=0.01 kg//m` `=(0.80xx10^(-6)m^(2))xx(12.5xx10^(3))=0.01 kg//m` speed of transverse waves produced in the string `v=sqrt((T)/(mu))=sqrt((64)/(0.01 kg//m))=80 m//s` the amplitude of the source is `a=1.0 cm` and the frequency is `f=20 Hz`. the angular frequency is `omega=2pif=40 pi//s`. also at `t=0`, the displacement is equal to its amplitute, `i.e.`, at `t=0, y=a`. the equation of motion of the source is, therefore, `y=(1.0 cm)cos[(40pis^(-1))t]` ...`(1)` the equation of the wave travelling on the string along the positive `X-`axis is obtained by replacing `t` by `[t-(x//v)]` in Eq. (i). It is, therefore, `y=(1.0 cm)cos[940pis^(-1)){t-(x//v)}x]` `=(1.0 cm)cos[(40pis^(-1))t-{(pi//2)m^-1)}x]` the displacement of the particle at `x=50 cm` at time `t=0.05 s` is obtained from Eq. (ii). `y=(1.0 cm)cos[(40pis^(-1))(0.05s)` `-{(pi//2)m^(-1}(0.5 m)]` `=(1.0 cm)cos[2pi-(pi//4)]` `=1.0 cm//sqrt(2)=0.71 cm` the velocity of the particle at position `x` at time `t` is also obtained from Eq. (ii). `V=(dely)/(delt)=-(1.0 cm)(40pi s^(-1))t-{(pi//2)m^(-1)}x]` `=-(40pi(cm)/(s))sin(2pi-(pi)/(4))` `=-(40pi)/sqrt(2) cm//s=-89 cm//s` |
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| 4. |
A `6.00m` segment of a long string has a mass of `180 g`. A high-speed photograph shows that the segment contlains four complete cycles of a wave. The string is vibrating sinusoidally with a frequency of `50.0 Hz` and a peak - to - valley dispalcement of `15.0 cm`. (The "peak-to-valley" displacment is the vertical distance from the furthest positive displacement to the farther negative displacement.) (a) Write the funcation that describe this wave traveling in the positive `x` direction. (b) Determine the average power being supplied to the string. |
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Answer» Correct Answer - (a) `y = (7.50 cm) sin ((4pi)/(3) xx -314t + phi)` (b) `(2025pi^(2))/(32) = 625 W` (a) `f = 500Hz` `mu = (180 xx 10^(-3))/(6) = 3 xx 10^(-2)Kg//m` `2A = 15 xx 10^(-2)m` `A = 7.5 xx 10^(-2)m` `4lambda = 6 , lambda = 3//2 m`. and `v = flambda = 75 m//sec` `omega = 2pif = 100pi` also `K = (omega)/(v) = (4pi)/(3)` If phase constant is `phi` then `:.` equation is `y = 7.5 xx 10^(-2) sin [100pit-(4pi)/(3)x + phi]` (b) `P_(av) = 2pi^(2)f^(2)A^(2)muv = (2025)/(32) pi^(2) = 625 w` |
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| 5. |
A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by `2.0 cm`. the motion of bar is continuous and is repeated regularly `125` times per second. If klthe distance between adjacent wave crests is observed to be `15.6 cm` and the wave is moving along positive `x-`direction, and at `t=0` the element of the string at `x=0` is at mean position `y =0` and is moving downward, the equation of the wave is best described byA. `y=(1 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`B. `y=(2 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`C. `y=(1 cm) cos [(40.3 rad//m) x-(786 rad//s)t]`D. `y=(2 cm) cos[(40.3 rad//m) x-(786 rad//s)t]` |
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Answer» Correct Answer - a Amplitude of wave, `A=(2.0cm)/(2)=1 cm` frequency of wave, `f=125 Hz` Wavelength of wave, `lambda=15.6 cm=0.156 m` Let equation of wave be, `y=A sin (kx-omega t+phi)` where `k=2pi//lambda=40.3 rad//m and omega=2pif=786 rad//s` Using initial conditions, `y(0,0)=0=A sin phi` and `(dely)/(delt)=Aomega cos philt0` we get, `phi=0` So, the equation of wave is `y=(1 cm) sin[(40.3 rad//m)x-(786 rad//s)t]` |
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| 6. |
Find speed of the wave generated in the spring as in the sitution shown. Assume that the tension is not affected by the mass of the cord. |
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Answer» `T = 20 xx 10 = 200 N` `v = sqrt((200)/(0.5)) = 20 m//s` |
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| 7. |
A metallic wire with tension `T` and at temperature `30^(@)C` vibrates with its fundamental frequency of `1 kHz`. The same wire with the same tension but at `10^(@)C` temperature vibrates with a fundamental frequency of `1.001 kHz`. The coefficient of linear expansion of the wire is equal to `10^(-K) .^(@)C`. Find `2K`. |
| Answer» Correct Answer - 8 | |
| 8. |
The average power transmitted through a given point on a string supporting a sine wave is 0.20 W when the amplitude of the wave is 2.0 mm. What power will be transmitted through this point if the amplitude is increased to 3.0 mm.A. `0.40` wattB. `0.80` wattC. `1.2` wattD. `1.6` watt |
| Answer» Correct Answer - D | |
| 9. |
A particle on a stretched string supporting a travelling wave, takess `5.0 ms` to move from its mean position to the extreme position. The distance between two consecutive particle, which are at their mean position, is `3.0 cm`. Find the wave speed `(in m//s)`. |
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Answer» Time period `T=4xx5 ms=20xx10^(-3)=2xx10^(-2)s` Frequency, `f=(1)/(T)=(1)/((2xx10^(--2)))=50 Hz` `lambda=2xx3 cm=6 cm` wave speed: `v=lambdaf=0.06xx50=3 m//s` |
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| 10. |
shows the position of a medium particle at t=0, supporting a simple harmonic wave travelling either along or opposite to the positive x-axis. (a) write down the equation of the curve. (b) find the angle `theta` made by the tangent at point P with the x-axis. (c ) If the particle at P has a velocity `v_(p) m//s`, in the negative y-direction, as shown in figure, then determine the speed and direction of the wave. (d) find the frequency of the wave. (e) find the displacement equation of the particle at the origin as a function of time. (f) find the displacement equation of the wave. |
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Answer» (a). From given curve: amplitude of the wave `A=10 cm` and wavelength `lambda=12 cm` `y=A sin(kx+-omegat+phi)=A sin((2pi)/(lambda)x+-(2pi)/(T)t+phi)`….(i) At `t=0, y=A sin(kx+phi)=10 sin(pi)/(6)x+phi)` At `x=0, y=-5 cm=10 sin phi` `sin phi=-(1)(2) impliesphi=-(pi)/(6)` hence `y=0.10(m)sin(pi)/(6)x-(pi)/(6))` (b). `tantheta =(dy)/(dx)=0.10xx(pi)(6) cos(pi)/(6)x-(pi)/(6))` At `x=6 m` `(dy)/(dx)tantheta=(pi)/(600cos(pi-(pi)/(6))=(pi)/(60)(-cos(pi)/(6))` `=-(pi)/(60).sqrt(3)/(2)=(-pisqrt(3))/(120)` `theta=tan^(-1)(-(pisqrt(3))/(120))` (c ). nParticle velocity `v_(p)=v(dy/dx)` `-v_(p)=-v(-pisqrt(3))/(120))` `v=(-120)/sqrt(3pi)v_(p)=(-40sqrt(3))/(pi)v_(p)` Along negative `x-`axis. (d). wave velocity `v=f lambda` `impliesf=(v)/(lambda0=(40sqrt(3)v_(p))/(pixx12)` `=(10sqrt(3))/(3pi)v_(p)s^(-1)` (e). hence displacement equation of medium particle at origin at `x=0`, from eq... (ii) `y=10(cm)sin[(20sqrt(3))/(3)v_(p)t-(pi)/(6)]` (f). As wave is travelling towards negative x-directin hence positive sign should be used between kx and `omegat` in eq. ..(i) The equation of the wave, `y=0.1(m)sin[(pi)/(6)x(m)+omegat-(pi)/(6)]` here `omega=2pi f=2pi(10sqrt(3))/(3pi)v_(p)=(20sqrt(3))/(3)v_(p)` Hence displacement equation of the wave `y=0.1(m)sin[(pi)/(6)x(m)+(20sqrt(3))/(3)v-(p)t-(pi)/(6)]`...(ii) |
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| 11. |
A simple harmonic plane wave propagatees along x-axis in a medium. The displacement of the particle as a function of time is shown in figure, for `x=0`(curve 1) and `x=7` (curve 2) The two particle are with a span of onewavelength. The speed of the wave isA. `12 m//s`B. `24 m//s`C. `8 m//s`D. `16 m//s` |
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Answer» Correct Answer - a Let general wave equation be `y=A sin(omegat-kx+phi)` `v=(dy)/(dt)= Aomega cos(omegat-kx+phi)` For curve `(1),x=0` at `t=0,x=0,` we have `y=0` `implies 0=A sin [phi]implies sin phi=0` `implies phi=0` or `pi` here `phi=pi` (because velocity is negative) for curve `(2),xx=7 cm` `t=0,x=7 cm, y=-1` `-1=sin(-kxx7+pi)` `implies sin(-7k+pi)=-1//2` `implies-7k+pi=2npi+(7pi)/(6)` or `2npi+(11pi)/(6)` here `implies-7k+pi=2npi+(11pi)/(6)` (because at `t=0`, velocity is positive) `implies-7((2pi)/(lambda))=2npi+(5pi)/(6)` implieslammbda=(-14pi)/((5pi)/(6)+2npi)` ` `implieslambda=(-84)/(12n+5)` for `n=-1,lambda=12 cm` for `n=-2,lambda=(84)/(19)cm` (not possible) because `lambdagt7 cm` `v=flambda=100xx(12)/(100)=12 m//s` |
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| 12. |
A simple harmonic plane wave propagatees along x-axis in a medium. The displacement of the particle as a function of time is shown in figure, for `x=0`(curve 1) and `x=7` (curve 2) The two particle are with a span of onewavelength. The wavelength of the wave isA. `6 cm`B. `24 cm`C. `12 cm`D. `16 cm` |
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Answer» Correct Answer - c Let general wave equation be `y=A sin(omegat-kx+phi)` `v=(dy)/(dt)= Aomega cos(omegat-kx+phi)` For curve `(1),x=0` at `t=0,x=0,` we have `y=0` `implies 0=A sin [phi]implies sin phi=0` `implies phi=0` or `pi` here `phi=pi` (because velocity is negative) for curve `(2),xx=7 cm` `t=0,x=7 cm, y=-1` `-1=sin(-kxx7+pi)` `implies sin(-7k+pi)=-1//2` `implies-7k+pi=2npi+(7pi)/(6)` or `2npi+(11pi)/(6)` here `implies-7k+pi=2npi+(11pi)/(6)` (because at `t=0`, velocity is positive) `implies-7((2pi)/(lambda))=2npi+(5pi)/(6)` implieslammbda=(-14pi)/((5pi)/(6)+2npi)` ` `implieslambda=(-84)/(12n+5)` for `n=-1,lambda=12 cm` for `n=-2,lambda=(84)/(19)cm` (not possible) because `lambdagt7 cm` `v=flambda=100xx(12)/(100)=12 m//s` |
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| 13. |
A plane undamped harmonic wave progates in a medium. Find the mean space density of energy becomes equal to `W_(0)` at an instant `t=t(0)+T//6`, where `t_(0)` is the instant when amplitude is maximum at this location and T is the time period of oscillation. |
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Answer» Let us consider the wave `Y=A cos(omegat-kx)`, then its energy density (energy per unit volume)is given by `W=pA^(2)omega^(2) sin ^(2) (omegat-kx)` `[`where `p` is of medium, for string waves, `mu =pS]`. Let us consider `x=0`, `t_(0)=0` at which amplitude is maximum At `t=t_(0)+(T)/(6)` And `t=A cos[(omegaT)/(6)]` and the energy density is `w=pA^(2)omega^(2)sin^(2)[(omegaT)/(6)]=pA^(2)omega^(2)sin^(2)(pi)/(3)` `W=(pA^(2)omega^(2))(3)/(4)` From given data, `W=W_(0)` `impliespA^(2)omega^(2)=(4)/(3)W_(0)=(2W_(0)/(3)` |
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| 14. |
The figure below is a representation of a simple harmonic progressive in the negative X-axis, at a given instant. The direction of the velocity of the particle at stage P in the figure is best represent by the arrow. A. `vec(P)A`B. `vec(P)B`C. `vec(P)C`D. `vec(P)D` |
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Answer» Correct Answer - a `v_(p)=-v((dy)/(dx))` here c is negative and `dy//dx` is positive at point `P`. Hence `V_(p)`is positive. So `P` is moving upward. |
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| 15. |
A simple harmonic progressive wave in a gas has a particle displacement of y=a at time `t=T//4` at the orgin of the wave and a particle velocity of y =v at the same instant but at a distance `x=lambda//4` from the orgin where T and `lambda` are the periodic time and wavelength of the wave respectively. then for this wave.A. the amplitude `A` of the wave is `A=2a`B. the amplitude `A` of the wave is `A=a`C. the equation of the wave can be represented by `y asin(v)/(a)[(t)-(x)/(V)]`D. The equation of the wave can be represented by `y 2a cos(v)/(a)[t-(x)/(V)]` |
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Answer» Correct Answer - b.,c. Let the equation to the wave be `y=A sin[2pi((t)/(T)-(x)/(lambda))+phi]` where `A` is the amplitude of the wave and `phi`, phase angle. It is given that `y=a`, when `x=0` and `t=T//4` and also that `y=v when `x=lambda//4` and `t=T//4` substituting in (i), `y=a=A sin((pi)/(2)+phi)` `y=v=(2piA)/(T)cos[2pi((1)/(4)-(1)/(4) )+phi]` `v=(2piA)/(T)cosphi` putting `phi=0, y=a=A`, so that amplitude `A=a` Also, `v=(2piA)/(T)[cos0]=(2pia)/(T)` `(2pi)/(T)=(v)/(a)` Hence the equation to the wave is `y=a sin(v)/(a)[t-(Tx)/(lambda)]` ` `y=a sin(v)/(a)[t-(x)/(V)]` where `V=(lambda)/(T)` is the velocity of the wave in the gas. |
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| 16. |
The same progressive wave is represented by two graphs I and II. Graph I shows how the displacement `y` varies with the distance x along the wave at a given time. Graph II shows how y varies with time t at a given point on the wave. The ratio of measurements AB to CD, marked on the curves represents:A. wave number kB. wave speed VC. frequency nD. angular frequency `omega` |
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Answer» Correct Answer - b `(measure AB)/(measure CD)=(lambda)/(T)=(1//K)/(1//W)=V_(Omega)` |
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| 17. |
A transverse periodic wave on a string with a linear mass density of 0.200 kg/m is described by the following equation `y=0.05 sin (420t-21.0 x)` where x and y are in metres and t is in seconds.The tension in the string is equal to :A. 32 NB. 42 NC. 66 ND. 80 N |
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Answer» Correct Answer - D |
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| 18. |
A transverse periodic wave on a string with a linear mass density of `0.200 kg//m` is described by the following equation `y = 0.05 sin(420t - 21.0 x)` where `x` and `y` are in meters and `t` is in seconds. The tension in the string is equal to :A. `32 N`B. `42 N`C. `66 N`D. `80 N` |
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Answer» Correct Answer - D `V_(omega) = (omega)/(k) = (420)/(21) = 20 :. V = sqrt((T)/(mu)) = 20 rArr T = (20)^(2)mu = 20^(2) xx 0.2 = 80N` |
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| 19. |
Graph shows three waves that are separately sent along a string that is stretched under a certain tension along x-axis. If `omega_(1),omega_(2) and omega_(3)` are their angular frequencies, respectively, then: A. `omega_(1)=omega_(3)gtomega_(2)`B. `omega_(1)gtomega_(2)gtomega_(3)`C. `omega_(2)gtomega_(1)=omega_(3)`D. `omega_(1)=omega_(2)=omega_(3)` |
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Answer» Correct Answer - a As `f _(1)=f_(2),f_(2)=(f)/(2),f_(3)=f` `:.Omega_(1)=2pifimplies Omega_(3)=2pif and Omega_(2)=pif` |
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| 20. |
A narrow pulse (for example, a short pip by a whistle) is sent across a medium. If the pulse rate is 1 after every 20 s (that is the whistle is blow 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» The frequency of the note produced by the whistle is not equal to `1//20` or `0.05 Hz`, it is only the frequency of pulse repetition. | |
| 21. |
A transverse harmonic wave of amplitude 0.01 m is genrated at one end (x=0) of a long horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time the displacement of the particle at x=0.1 m is -0.005 m and that of the particle at x=0.2 m is +0.005 m. calculate the wavelength and the wave velocity. obtain the equetion of the wave assuming that the wave is traveling along the + x-direction and that the end x=0 is at he equilibrium position at t=0. |
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Answer» Since the wave is travelling along `+ x-`direction and the displacement of the end `x=0` is at time `t=0`, the general equation of this wave is `y(x,t)=A sin{(2pi)/(lambda)(vt-x_(1))}`...`(i)` where `A=0.01 m` when `x=0.1 m, y=-0.005 m` Putting all values, we get `-0.005=0.01 sin{(2pi)/(lambda)(vt-x_(1))}` where `x_(1)=0.1 m` or, `sin{(2pi)/(lambda)(vt-x_(1)}=-(1)/(2)` `:.` Phase sintheta`_(1)=(2pi)/(lambda)(vt-x_(1))=(7pi)/(6)`....`(ii)` When `x=0.2 m y=+0.005`. therefore, we have `+0.005=0.01 sin{(2pi)/(lambda)(vt-x_(2))}` where `x_(2)=0.2 m` `theta_(2)=(2pi)/(lambda)(vt-x_(2))=(pi)/(6)` ......`(iii)` From Eqs.(ii)and (iii) `Deltatheta=theta_(1)-theta_(2)=pi` now, `Deltatheta=(2pi)/(lambda)Deltax` Thus, `pi=-(2pi)/(lambda)(x_(1)-xx_(2))=(-2pi)/(lambda)(0.1-0.2)` or `lambda=0.2 m` Now, frequency g og the wave `=`frequency of the tuning for `k=500 Hz`. hence, wave velocity `v=f lambda=500xx0.2=100 m//s` substituting for `A`,lambda and `v` in eq. (i) we get `y(x,t)=0.01 sin(10pi(100t-x))` this is the lequation of the wave where y and x are in metres and t in seconds. |
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| 22. |
A harmonic wave has been set up on a very long string which travels along the length of string. The wave has frequency of 50 Hz. Amplitude 1 cm and wavelength 0.5 m. for the above described wave. Statement (i): time taken by a point on the string to travel a distance of 8 m along the length of strime is 0.32 s. Statement (ii): time taken by a point in the string to travel a distance of 8m, once the wave has reached at that point and sets it into motion is 0.32 s.A. Both the statement are correctB. statement I is correct but statement II is incorectC. statement I is incorrect but statement II is correctD. both the statement are incorrect |
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Answer» Correct Answer - b The wave is travelling along the length of a string while particles constituting the string are oscillation in direction perpendicular to the length to string. In one time period (cycle), the wave moves forward by one wavelength while the particle on string travels a distance of 4 times the amplitude. Here, `T=1//f=0.02 s` Wave speed, `v=flambda=25 m//s` Time taken by wave to travel a distance of 8 m, `t_(1)=8//25 s=0.32 s`. Time taken by particle on string to travel a distance of 8 m. `t_(2)=(8xxT)/(4 "times amplitude") =(8)/(4xx0.01)xx0.02=4 s` |
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| 23. |
A simple harmonic oscillator at the point x=0 genrates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. the rope has a linear mass density of `50.0 g//m` and is strectches with a tension of 5.00 N. (a) determine the speed of the wave. (b) find the wavelength. (c ) write the wave function y(x,t) for the wave, Assume that the oscillator has its maximum upward displacement at time t=0. (d) find the maximum transverse acceleration at time t=0. (d) find the maximum transverse of points on the rope. (e) in the discussion of transverse waves in this chapter, the force of gravity was ignored. is that a reasonable assumption for this wave? explain. |
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Answer» (a). `V=sqrtF//mu=sqrt((5.00 N)//(0.0500 kg//m))=10.0 m//s` (b). `Lambda= c//f=)10.0 m//s)//(40.0 Hz)=0.250 m` (c ) `y(x,t)=A cos(kx-omegat) ("note":y(0.0)=A`, as specified.)` `k=2pi//lambda=8pi rad//m, omega=2pif=80.0pi rad//s` `y(x,t)=(300 cm)cos[pi(8.00 rad//m)x-(80.0pi rad//s)t]` (d). `v_(y)=+Aomegasin(kx-omegat) and a_(y)=-Aomega^(2)cos(kx-omegat)` `a_(y,max)=Aomega^(2)=A(2pif)^(2)=1890 m//s^(2)` (e) `a_(y,max)` is much larger than `g`, so `g` can be ignored. |
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| 24. |
A travelling wave is described by the equation `y = y_(0) sin ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity ifA. `lambda=(piy_(0))/(4)`B. `lambda=(piy_(0))/(2)`C. `lambda=piy_(0)`D. `lambda=2pi Y_(0)` |
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Answer» Correct Answer - b maximum particle velocity `= 4 `wave velocity `Aomega=4 flambda` `y_(0)2pi f=4flambda` `lambda=(piy_(0))/(2)` |
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| 25. |
A travelling wave is described by the equation `y = y_(0) sin ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity ifA. `lambda = piY_(0)//4`B. `lambda = piY_(0)//2`C. `lambda = piY_(0)`D. `lambda = 2pi Y_(0)` |
| Answer» Correct Answer - B | |
| 26. |
A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequancy 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) calculate the average power carried by the wave. (b) what happenes to the avarage power if the wave amplitude is halved? |
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Answer» (a). As we know `P_(ave)=(1)/(2)(sqrtmuF)omega^(2)A^(2)` =(1)/(2)[sqrt{(3.00xx10^(-3) kg)/(0.80 m)(25.0 N))][(2pi(120.0 Hz)]^(2)(1.6xx10^(-3))^(2)` `=0.223 W` (b). Halving the amplitude quarters the average power, so `0.056 W`. |
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| 27. |
Small amplitude progressive wave in a stretched string has a speed of `100 cm//s`.and frequency 100 Hz. The phase difference between two points 2.75 cm apart on the string in radians, is |
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Answer» Correct Answer - b `v=1 m//s, v=100 Hz` `lambda=(v)/(v)=(1)/(100)m=1 cm` `Deltaphi=(2pi)/(lambda)xx2.75 rad=5.5 pi rad=(11pi)/(2) rad` |
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| 28. |
The diagram below shows an instantaneous position of a string as a transverse progressive wave travels along it from left to right Which one of the following correctly shows the direction of the velocity of the points 1,2 and 3 on the string?A. `rarr" " rarr" " rarr`B. `larr" " larr" " rarr`C. `darr" " darr" " darr`D. `darr" " uarr" " darr` |
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Answer» Correct Answer - C |
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| 29. |
For the wave shown in figure, write the equation of this wave if its position is shown at `t= 0`. Speed of wave is `v = 300m//s`. A. `y=(0.06 m)cos[(78.5 m^(-1))x+(23562 s^(-1))t]m`B. `y=(0.06 m)sin[(78.5 m^(-1))x-(23562 s^(-1))t]m`C. `y=(0.06 m)sin[(78.5 m^(-1))x+(23562 s^(-1))t]m`D. `y=(0.06 m)cos[(78.5 m^(-1))x-(23562 s^(-1))t]m` |
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Answer» Correct Answer - b The amplitude, `A=0.06 m` `(5)/(2)lambda=0.2 m` `:. Lambda=0.08 m` `f=(v)/(lambda)=(300)/(0.08)=3750 Hz` `k=(2pi)/(lambda)=78.5 m^(-1)` and `omega=2pi f=23562 rad//s` At `T=0,x=0, (dy)/(dx)=`positive and the given curve is a sine curve. Hence, equation of wave travelling in positive `x-`direction should have the form, `y(x,t)=A sin (kx-omegat)` substituting the values, we have `y=(0.06 m) sin [(78.5 m^(-1))x-(23562 s^(-1))t] m` |
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| 30. |
Three waves of equal frequency having amplitudes `10mum`, `4mum`, `7mum` arrive at a given point with successive phase difference of `pi//2`, the amplitude of the resulting wave in `mum` is given byA. `7`B. `6`C. `5`D. `4` |
| Answer» Correct Answer - C | |
| 31. |
Assertion: In a small segment of string carrying sinusoidal wave, total energy is conserved. Reason: Every small part moves in SHM and total energy of SHM is conserved.A. Statement I is true, statement II is true and statement II is the correct explaination for statement I.B. Statement I is true, statement II is true and statement II is NOT the correct explaination for statement I.C. Statement I is true, statement II is false.D. Statement I is false, statement II is true. |
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Answer» Correct Answer - d Every small segament is acted upon by forces from both sides of it hence enrgy is not conserved, rather it is transmitted by the element. |
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| 32. |
A wave pulse is generated in a string that lies along `x-`axis. At the points `A` and `B`, as shown in figure, if `R_(A)` and `R_(B)` are ratio of magnitudes of wave speed to the particle speed, then A. `R_(A)gtR_(B)`B. `R_(B)gtR_(A)`C. `R_(B)=R_(A)`D. information is not sufficient |
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Answer» Correct Answer - a Slope at any point on the string in wave motion represents the ratio of particle speed to wave speed. Therefore, slope Bltslope A Hence `R_(A)gtR_(B)`. |
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| 33. |
A transverse sinusoidal wave is generted at one end of long, horizontal string by a bar that moves up and down through a distance of `1.00 cm`. The motion is continuous and is repreated regularly `120` times per second. The string has linear density `90 gm//m` and is kept under a tension of `900 N`. Find : What is the transverse displacement `y`(in `cm`) when the minimum power transfer occurs [Leave the answer in terms of `pi` wherever it occurs]A. `0`B. `(1)/(2)`C. `(1)/(4)`D. `1` |
| Answer» Correct Answer - B | |
| 34. |
The displacement of partcles in a string streched in the x-direction is by `y`. Among the following expressions for `y`, those describing wave motion are :A. `cos (kx) sin(omegat)`B. `k^(2)x^(2) - omega^(2)t^(2)`C. `cos^(2)(kx + omegat)`D. `cos(k^(2)x^(2) - omega^(2)t^(2))` |
| Answer» Correct Answer - A::C | |
| 35. |
A sinusoidal wave is traveling on a string with speed 40 cm/s. The displacement of the particle of the string at x = 10 cm varies with time according to y = (5.0cm) sin`(1.0-4.0s^(-1))t)`The linear density of the string is 4.0 g/cmA. the frequency of wave is 0.64 HzB. the wavelenght of the wave 63 cmC. the tension in the string 0.064 ND. the wave equation is y = (5.0cm) sin `(0.1x-(4.0s^(-1)t)` |
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Answer» Correct Answer - A::B::C::D |
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| 36. |
The linear density of a vibrating string is `10^(-4) kg//m`. A transverse wave is propagating on the string, which is described by the equation `y=0.02 sin (x+30t)`, where x and y are in metres and time t in seconds. Then tension in the string isA. 0.09 NB. 0.36 NC. 0.9 ND. 3.6 N |
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Answer» Correct Answer - a `y=0.02 sin (x+30t)` comparing with standard equation `y=A sin (Kx+omegat),omega=30, K=1` velocity of wave, `v=(omega)/(K)=(30)/(1)=30 m//s` Expression `v=sqrt((T)/(m))` gives Tension `T=v^(2)m=(30)^(2)xx10^(-4)` `=0.09 N` |
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| 37. |
Two small boats are `10m` apart on a lake. Each pops up and down with a period of 4.0 seconds due to wave motion on the surface of water. When one boat is at its highest point, the other boat is its lowest point. Both boats are always within a single cycle of the waves. The speed of the waves is:A. `2.5 m//s`B. `5.0 m//s`C. `14 m//s`D. `40 m//s` |
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Answer» Correct Answer - B Distance between boat `= (lambda)/(2) = 10 m` `rArr = 20m` time penod, `T = 4` sec . `:. V = lambda//T = 20 m //4 sec = 5 m//s`. |
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| 38. |
A transverse wave is propagating along `+x` direction. At `t=2` sec the particle at `x=4`m is at `y=2` mm. With the passage of time its `y` coordinate increases and reaches to a maximum of 4 mm. The wave equation is (using `omega` and `k` with their usual meanings)A. `y=4sin(omega(t+2)+k(x-2)+(pi)/(6))`B. `y=4sin(omega(t+2)+k(x)+(pi)/(6))`C. `y=4sin(omega(t-2)-k(x-4)+(5pi)/(6))`D. `y=4sin(omega(t-2)-k(x-4)+(5pi)/(6))` |
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Answer» Correct Answer - D |
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| 39. |
A sinusoidal wave is propagating along a streched string that lies along the x-axis. The displacement of the string as a function of time is graphed in for particles at x-0 and at x=0.0900 m. (a) what is the amplitude of the wave? (b)what is the period of the wave? (c ) you are told that the two points x=0 and x=0.0900 m are within one wavelength of each other. if the wave is moving in the +x-direction, determine the wavelength and the wave speed. (d) if instead the wave is moving in the -x-direction, determine the wavelength and the wave speed. (e) would it be possible to derermine definitely the wavelength in parts (c ) and (d) if you were not told that the two points were within one wavelength of each other? why ot why not? |
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Answer» Let equation of wave is `y=A sin (omegat-kx+theta)` `V_(p)=(dy)/(dt)=Aomegacos(omegat-kx+theta)` `atx=0,t=0`,we have `y =0,v_(p)gt0` `implies0= A sin thetaimplies theta=0` or `pi` But `v_(p)gt0`,so `theta=0` at `t=0, x=0.090 m` `y=A=4 mmrarr amplitude` `A=A sin (-k(0.090))` `implies -k(0.090)=2npi+(pi)/(2)` `implies -(2pi)/(lambda)(0.090)=2npi+(pi)/(2)` `implies lambda=(-0.36)/(4n+1)` for `n=-1,lambda=0.12 m` from graph, time period `=T=0.04 s` wave speed: `v=(lambda)/(T)=(0.12)/(0.04)=3 m//s` if wave is moving in negativex-direction: let wave equation is `y=A sin(omegat+kx+theta)` Processing as above, we will get `lambda=(0.36)/(4n+1)` `for `n=0,lambda=0.36 m` wave speed: `v=(lambda)/(T)=(0.36)/(0.04)=9 m//s` |
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| 40. |
A wave disturbance in a medium is described by `y(x, t) = 0.02 cos(50pit + (pi)/(2))cos(10pix)` where `x and y` are in meter and `t` is in second`A. A node occurs at `x = 0.15 m`B. An antinode occurs at `x = 0.3 m`C. The speed pf wave is `5 ms^(-1)`D. The wavelength is `0.2 m`. |
| Answer» Correct Answer - A::B::C::D | |
| 41. |
A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at `x=2` cm is found at its mean position and it is moving in positive y-direction at `t=1` s. the amplitude of the wave, the wavelength and the angular frequency of the wave are `0.1m,pi//4m` and `4pi rad//s`, respectively. The instantaneous power transfer through `x=2 m` and `t=1.125 s` isA. `10 J//s`B. `4pi//3 J//s`C. `2pi//3 J//s`D. 0 |
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Answer» Correct Answer - d Time period of oscillation `T=(2pi)/(omega)=(2pi)/(4pi)=(1)/(2)s` Hence at `t=1.125 s`, that is, at `T//4` seconds after `t=1 s`, the particle is at extreme position. Hence, instantaneous power at `x=2` at `t=1.125 s` is zero. |
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| 42. |
A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at `x=2` cm is found at its mean position and it is moving in positive y-direction at `t=1` s. the amplitude of the wave, the wavelength and the angular frequency of the wave are `0.1m,pi//4m` and `4pi rad//s`, respectively. The equation of the wave isA. `y=0.1 sin(4pi(t-1)+8(x-2))`B. `y=0.1 sin((t-1)-(x-2))`C. `y=0.1 sin(4pi(t-1)-8(x-2))`D. none of these |
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Answer» Correct Answer - a The equation of wave moving in negative `x`-direction, assuming origin of position at `x=2` and origin of time (i.e., initial time) at `t=1 s`. `y=0.1 sin(4pi t+8x)` shifting the origin of position to left by `2 m`, to `x=0`. Also shifting the origin of time backwards by `1 s`,that is to `t=0 s`. `y=0.1 [4pi(t-1)+8(x-2)]` |
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| 43. |
A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at `x=2` cm is found at its mean position and it is moving in positive y-direction at `t=1` s. the amplitude of the wave, the wavelength and the angular frequency of the wave are `0.1m,pi//4m` and `4pi rad//s`, respectively. The speed of particle at `x=2 m` and `t=1 s` isA. `0.2pi m//s`B. `0.6pi m//s`C. `0.4pi m//s`D. 0 |
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Answer» Correct Answer - c As given the particle at `x-2`is at mean position at `t=1 s`. :. Its velocity `v=omegaA=4pixx0.1=0.4pi m//s`. |
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| 44. |
The displacement of a wave disturbance propagating in the positive x-direction is given by `y =(1)/(1 + x^(2))`at `t = 0` and `y =(1)/(1 +(x - 1)^(2))` at `t =2s` where, `x` and `y` are in meter. The shape of the wave disturbance does not change during the propagation. what is the velocity of the wave?A. `1 m//s`B. `1.5 m//s`C. `0.5 m//s`D. `2 m//s` |
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Answer» Correct Answer - c At t=2 s `y=(1)/([1+(x-1)^(2)])` or `x-vt=x-1 implies 1=vt` `implies1=vxx2` `impliesv=0.5 m//s` |
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| 45. |
A transverse wave of amplitude `0.5 m` and wavelength 1 m and frequency 2 Hz is propagating in a string in the negative x - direction. The expression for this wave is [A. `y(x,t) = 0.5 sin (2pix - 4pit)`B. `y(x,t) = 0.5 cos (2pix - 4pit)`C. `y(x,t) = 0.5 sin (pix - 2pit)`D. `y(x,t) = 0.5 cos (2pix - 2pit)` |
| Answer» Correct Answer - B | |
| 46. |
The amplitude of a wave disturbance propagating along positive X-axis is given by `=1/(1+x^(2))` at t=0 and `y=1/[1+(x-2)^(2)]` at t=4 s where x and y are in metre. The shape of wave diturbance does not change with time. The velocity of the wave isA. `0.5 m//s`B. `1 m//s`C. `2 cm/s`D. `4 m//s` |
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Answer» Correct Answer - a `y=(1)/(1+(x^(2))` at `t=0` and `y=(1)/(1+(x-2)^(2))` at `t=4s` `v=(Deltax)/(Deltat)=(x-(x-2))/(4-0)=(2)/(4)=0.5 m//s` |
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| 47. |
If the speed of the wave shown in the figure is `330m//s` in the given medium then the equation of the wave propagating in the positive x-direction will be (all quantities are in M.K.S units)A. `y=0.05 sin 2pi(4000t-12.5x)`B. `y=0.05sin2pi(4000t-122.5x)`C. `y=0.05 sin 2pi(3300t-10x)`D. `y=0.05 sin 2pi(3300t-10t)` |
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Answer» Correct Answer - c |
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| 48. |
A transverse harmonic wave on a strin is decribed by `y(x,t) = 3.0 sin (36 t + 0.018x + pi//4)` Where `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 starting point ? What is the least distance between two successive crests in the wave ? |
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Answer» Correct Answer - (a) A travelling wave. It travels from right to left with a speed of `20 ms^(-1)`. (b) `3.0 cm, 5.7 Hz` (c) `pi//4` , (d) `3.5 m` |
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| 49. |
For brass, bulk modulus and modulus of rigidity are `10.72 xx 10^(10)Nm^(-2)` and `3.6 xx 10^(10)Nm^(-2)`. If the density of brass is `8.4 xx 10^(3) kgm^(-3)`, find the speed of longitudinal and transverse waves in it. |
| Answer» Correct Answer - `4.3 xx 10^(3)ms^(-1); 2.07 xx 10^(3)ms^(-1)` | |
| 50. |
A wire having a linear density of `0.05 gm//c` is stretched between two rigid supports with a tension supports with a tension of `4.5 xx 10^(7)` dynes. It is oberserved that the wire resonates at a frequency of `420 "cycles"//"sec"`. the next higher frequency at which the same wire resonates is `490 "cycles"//"sec"`. Find the possible mode of vibration of string initially : |
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Answer» Correct Answer - 6 `V = sqrt((T)/(mu)) = sqrt((4.5 xx 10^(7))/(0.05))` `(n)/(2l) sqrt((4.5 xx 10^(7))/(0.05)) = 420 "___"(1)` `(n + 1)/(2l) sqrt((4..5 xx 10^(7))/(0.05)) = 490 "__"(2)` From equation `(1)` & `(2) rArr (n)/(n + 1) = (6)/(7) rArr n = 6` Put `n` in `(1) :. (6)/(2l) 3 xx 10^(2) = 420` `l = (30000)/(140) , l = (1500)/(7) = 214 cm` |
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