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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.
| 201. |
Assertion `:` A simple harmonic motion of amplitude `A` has a time period `T`. The acceleration of the oscillator when its displacement is half the emplitude is `2pi^(2)A//T^(2).` Reason`:` In S.H.M., acceleration is directly proportional to the displacementA. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - a Here, both assertion and reason are correct and reason is the correct explanation of assertion. ltbr. Acceleration, `a=-omega^(2)y=-omega^(2)A//2` `=-(4pi^(2))/(T^(2))xx(A)/(2)=-(2pi^(2)A)/(T^(2))` |
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| 202. |
Does a vibrating body always produce sound? |
| Answer» No. Sound is produced only when frequency of vibrating lies between 20Hz and 20kHz. | |
| 203. |
A travelling harmonic wave on a string is described by `y(x,t)=7.5sin(0.0050x +12t_pi//4)` (a) what are the displacement and velocity of oscillation of a point at `x=1cm, ` and `t=1s` ? I sthis velocity equal to the velocity of wave propagation ? (b) Locate the point of the string which ahve the same transverse displacement and velocity as `x=1cm` point at `t=2s, 5s and ` 11s. |
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Answer» The travelling harmonic wave is `y(x,t)=7.5sin (0.0050x +12t+pi//4)` At `x=1cm `and `t=1 sec, y(1,1)=7.5sin(0.005xx1+12xx1+pi//4)=7.5sin(12.005+pi//4)` …(i) Now, `theta=(12.005+pi//4) ` radian `=(180)/(pi)(12.005+pi//4)degree=(12.005xx180)/((22)/(7))=732.55^(@)` `:.` from (i), `y(1,1)=7.5 sin (732.55^(@))=7.5sin(720+12.55^(@))` `=7.5sin12.55^(@)=7.5xx0.2173=1.63cm` velocity of oscillation, `upsilon(d)/(dt)[y(1,1)}=(d)/(dt)[7.5sin(0.005x+12t+(pi)/(4))]` `=7.5xx12cos[0.005+12t+(pi)/(4)]` At, `x=1cm, t=1sec. ` `upsilon=7.5xx12cos(0.005+12+pi//4)=90cos(732.55^(@))=90cos(720+12.55^(@))` `upsilon=90cos(12.55^(@))=90xx0.9765=87.89cm//s`. Comparing the given eqn. with the standard form `y(x,t)=rsin[(2pi)/(lambda)(upsilont+x)+phi_(0)]` we get, `r=7.5, (2piupsilon)/(lambda)=2piv=12 or v=(6)/(pi)` `(2pi)/(lambda)=0.005. ` `:. lambda=(2pi)/(0.005)=(2xx3.14)/(0.005)=1256cm=12.56m` velocity of wave propagtion, `upsilon=vlambda=(6)/(pi)xx12.56=24m//s` We find that velocity at `x=1cm, t=1` sec is not equal to velocity of wave propagation. (b) Now, all points which are at a distance of `+-lambda, +-2lambda, +-3lambda` from `x =1`cm will have same transverse displacement and velocity . As `lambda=12.56m` therefore, all points at distance `+-12.6m, +-25.2m, +-37.8m, ...` from `x=1cm` will have same displacement and velocity, `x=1cm` point `t=2s`, 5s and 11s. |
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| 204. |
A particle moving in a straight line has velovity v given by `v^(2)=alpha-betay^(2)` , where `alpha` and `beta` are constants and y is its distance from a ficed point in the line. Show that the motion of particle is SHM. Find its itme period and amplitude. |
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Answer» Given `v^(2)=alpha-betay^(2)` …(i) Differentiating it w.r.t. time t, we have, `2v(dv)/(dt)=-beta2y(dy)/(dt)=-beta2(v)` `:. (dv)/(dt)=-betay` …(ii) It means, acceleration, `A=(dv)/(dt)=-betay` As, `Apropy` and `-ve` sign shows that acceleration is directed towards mean position, so if the particle is left free, it will execute linear S.H.M. Here, `omega^(2)=beta or omega=sqrt(beta)` `:.` time period, `T=(2pi)/(omega)=(2pi)/(sqrt(beta))` We know that, `v=0, ` when `y=r`. From (i), `0=alpha-betar^(2) or r=sqrt((alpha)/(beta)), i.e., ` amplitude `=r=sqrt((alpha)/(beta))` |
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| 205. |
When an air column at `27^(@)C` and a tuning fork are sounded together, 5 beats per second are produced. The frequency of the fork is less than that of air column. No beat is heard at `-3^(@)C`. Determine the frequency of the fork. |
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Answer» Let the frequency of tuninig fork be v. Therefore, frequency of air column at `27^(@)C=(v+5)` As no beat is heard at `-3^(@)C`, therefore, frequency of air column at `-3^(@)C=v` `:. (v+5)/(v)=(upsilon_(26^(@)C))/(upsilon_(-3^(@)C))` `=sqrt((273+27)/(273-3))=sqrt((300)/(270))` `1+(5)/(v)=(1+(1)/(9))^(1//2)=1+(1)/(18)` `v=5xx18=90Hz` |
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| 206. |
In the given progressive wave `y=5sin (100pit-0.4pix)` where y and x are in m, t in s, what is the (a) amplitude (b) wavelength (c) frequency (d) wave velocity (e) particle velocity amplitude. |
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Answer» The given equation is `y=5sin(100pit-0.4pix)` Compart is with the standard form of eqn. of progressive wave `:y=rsin[(2pit)/(T)-(2pi)/(lambda)x]` (a) amplitude, `r=5m` (b) `(2pi)/(lambda)=0.4pi, lambda=(2pi)/(0.4pi)=5m` (c) `(2pi)/(T)=100pi, (1)/(T)=v=(100pi)/(2pi)=50Hz` (d) wave velocity, `upsilon=vlambda=50xx5=250m//s` (e) particle velocity, `(dy)/(dt)=5xx100picos(100pit-0.4pix)` `((dy)/(dt))_(max)=500pixx1m//s=500pim//s` |
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| 207. |
A column of air at `51^(@) C` and a tuning fork produce `4` beats per second when sounded together. As the temperature of the air column is decreased, the number of beats per second tends to decrease and when the temperature is `16^(@) C` the two produce `1` beat per second. Find the frequency of the tuning fork. |
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Answer» Let the frequency of tuning fork be v. As `v=(upsilon)/(lambda)`, therefore, with decrease in temperature, `upsilon` decreases and hence v decreases. As number of beats `//` sec decreases with decrease in frequency of air column, therefore, frequency of air column is greater than the frequency of funing fork. `:.` Frequency of air column at `51^(@)C=(V+4),` and frequency of air column at `16^(@)C=(v+1)` Now, `(v+4)/(v+1)=(upsilon_(51))/(upsilon_(16))=sqrt((273+51)/(273+16))=sqrt((324)/(289))=(18)/(17)` `18v+18=17v+68` `v=68-18=50Hz` |
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| 208. |
A tuning fork of frequency 220Hz produces sound waves of wavelength 1.5m in air at NTP. Calculate the increase ini wavelenth, when temperature of air in `27^(@)C`. |
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Answer» Here, `V=220Hz, lambda_(1)=1.5m, T_(1)-0^(@)C=273K, upsilon_(1)=vlambda_(1)=220xx1.5=330m//s` `T_(2)=27^(@)C=27+273=300K.` `upsilon_(2)=upsilon_(1)sqrt((T_(2))/(T_(1)))=330sqrt((300)/(273))=345.9m//s` `lambda_(2)=(upsilon_(2))/(n)=(345.9)/(220)=1.57m` Increase in wavelength `=lambda_(2)-lambda_(1)=1.57-1.50=0.07m` |
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| 209. |
A tuning fork produces 4 beats`//`s when sounded with a fork of frequency 512 Hz. The same tuning fork. When sounded with another fork of frequency 514 Hz produces 6 beats `//` second. Find the unknown frequency of the fork. |
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Answer» Let the unknown frequency be v. `:. v=512+-4=516 or 508 Hz` Also,` v=514+-6=520 or 508 Hz.` As common frequency is 508 Hz. This must be the unknown frequency. |
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| 210. |
Write the equation of a progressive wave propagating along the positive x-direction, whose amplitude is 5cm, frequency 250Hz and velocity 500 m`//` s. |
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Answer» Here, `r=5cm=0.05m, v=250Hz,` `upsilon=500m//s` `T=(1)/(v)=(1)/(250)s, lambda=(upsilon)/(v)=(500)/(250)=2m` The equation of the wave is `y=rsin2pi((t)/(T)-(x)/(lambda))` `y=0.05sinpi(500t-s)metre` |
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| 211. |
A tuning fork of frequency 220Hz produces sound waves of wavelength 1.5m in air at NTP. Calculate the increase in wavelength, when temperature of air in `27^(@)C`. |
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Answer» Here,`v=220Hz. lambda _(0)=1.5m.` Speed of sound at N.T.P, `v_(0)=vlambda_(0)=220xx1.5=330m//s` Now, `T=27^(@)C=(27+273)K=300K` If v is speed of sound in air at 300K. Then `v=v_(0)sqrt((T)/(T_(0)))=330xxsqrt((300)/(273))` `=330xx1.048=345.93m//s` wavelength of sound waves at 300K. `lambda=(upsilon)/(v)=(345.93)/(220)=1.57m` `:.` Increase in wavelength `=lambda-lambda_(0)` `=1.57-1.5=0.07m` |
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| 212. |
Speed of sound waves in airA. is independent of temperatureB. increases with pressureC. increases with increase in humidityD. decreases with increase in humidity. |
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Answer» Correct Answer - C Speed of sound waves in air increases with in humidity. This is because presence of moisture decreases the density of air. |
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| 213. |
Light waves can be polarised but sound waves cannot be. Why? |
| Answer» Polarisation is the property shown by transverse waves. Sound waves are longitudinal and light waves are transverse. | |
| 214. |
If the splash is hear 4.23 seconds after a stone is dropped into a well, 78.4 metres deep, find the velocity of sound in air. |
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Answer» Here, depth of the well, `s=78.4metres` Total time after which splash is heard `=4.23s`, If `t_(1)=` time taken by stone to hit the water surface in the well. `t_(2)=` time taken by splash of sound to reach the top of the well. then, `t_(1)+t_(2)=4.23s` ...(i) Now, for downwards journey of stone, `u=0, a=9.8ms^(-2), s=78.4m, t=t_(1)=?` As `s=ut+(1)/(2)at^(2)` `:.78.4=0+(1)/(2)xx9.8t_(1)^(2)=4.9t_(1)^(2)` `t_(1)^(2)=(78.4)/(49.9)=16 or t_(1)=sqrt(16)=4s` From (i), `t_(2)=4.23-t_(1)=4.23-4=0.23` second If v is the velocity of sound in air,then `v=(distance (s))/(time(t_(2)))=(78.4)/(0.23)=34.87ms^(-1)` |
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| 215. |
Why is sound hear in `CO_(2)` more intense in comparison to sound heard in air? |
| Answer» This is because intensity of sound increases with increase in density of the medium. | |
| 216. |
A person deep inside water cannot hear sound produced in air. Why? |
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Answer» As speed of sound in water is roughly four times the speed of sound in air, therefore `mu=(sini)/(sinr)=(v_(a))/(v_(w))=(1)/(4)=0.25` For refraction, `r_(max)=90^(@)` `:. (sini)_(max)=0.25` `:. i_(max)=sin^(-1)(0.25)~~1.4^(@).` That is why most of sound produced in air and falling at `/_igt14^(@)` gets reflected in air only and person deep inside water cannot hear the sound. |
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| 217. |
A fork of frequency 400Hz is in unison with a sonometer wire. How many beats `//` sec will be hear when tension in the wire is increased by `2%` ?A. 2B. 4C. 1D. 3 |
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Answer» Correct Answer - B `4` |
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| 218. |
Two identical sinusoidal waves, moving in same direction along a stretched string, interfere, with each other. The amplitude of each wave is 10.0mm, and the phase difference between them is `80^(@)`. What is the amplitude of the resultant wave and the nature of interference ? |
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Answer» Here, amplitude of each wave `a=b=10mm` phase difference `=phi=80^(@)` The amplitude of resultant wave is, `R=sqrt(a^(2)+b^(2)+2abcosphi)` `=sqrt(10^(2)+10^(2)+2xx10xx10cos80^(@))` `=sqrt(100+100+200xx0.1736)` `=sqrt(234.72)=15.32mm` |
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| 219. |
Two sounds of very close frequencies 510Hz and 514Hz are produced simultaneously. What is the frequency of resultant sound? How many beats can be heard in 5sec? |
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Answer» Here, `v_(1)=510Hz, v_(2)=514Hz.` Resultant frequency, `v=(v_(1)+v_(2))/(2)=(510+514)/(2)=512Hz` No. of beats `//` sec `=514-510=4` Beats heard in 5sec `=4xx5=20` |
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| 220. |
The distance between two points on a stretched string is 20 cm. A pregressive wave of frequency 400 hertz travels on the string with a velocity of `100m//s`. Calculate phase differency between the points. |
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Answer» `lambda=(v)/(n)=(100)/(400)=0.25m` `Deltaphi=(2pi)/(lambda)Deltax=(2pi)/(0.25)xx((20)/(100))=1.6piradian` |
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| 221. |
Two sources of sound produce 20 beats in 4 second. What is the difference in frequencies of the two sources? |
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Answer» Here, no. of beats `//` sec, `m=(20)/(4)=5` `:. N_(1)-n_(2)=5` |
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| 222. |
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength. |
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Answer» The wave pattern shown in figure, represents a stationary wave. At `t=0, t=T//2` and `t=T` , points at x in cm, `0,10,20,30,40` are permanantly at rest. They represent the position of nodes. Distance between two successive nodes `=(lambda)/(2)=10cm.` `:. lambda=20cm` |
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| 223. |
Can beats be observed in two light sources of nearly equal frequencies ? |
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Answer» No, as emission of light from atoms is a random and rapid phenomenon. The phase at a point due to two independent light sources will change rapidly and randomly. Therefore, instead of beats, we shall get uniform intensity. However, if light sources are laser beams of nearly equal frequencies, we may be able to observe the phenomenon of beats in light. |
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| 224. |
Is it necessary for production of beats that the two waves must have exactly equal amplitudes? |
| Answer» No. It is not at all necessary. However, the beats become more distinct when the amplitude of two waves are exactly equal. | |
| 225. |
A string of length `0.4 m` and mass `10^(-2)kg` is tightly clamped at its ends. The tension in the string is `1.6 N`. Identical wave pulse are produced at one end at equal intervals of time, `Deltat`. The minimum value of `Deltat` which allows constructive interference of successive pulse isA. `0.05s`B. `0.10s`C. `0.20s`D. `0.40s` |
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Answer» Correct Answer - A `n=(1)/(2l)sqrt((T)/(m))=(1)/(2xx0.4)sqrt((1.6)/(10^(-2)//0.4))=10Hz. ` `:.` time period `t=1//n=(1)/(10)=0.1s.` As distance between two fixed ends `=lambda//2` `:.` for constructive interference, minimum value of time `Deltat=(t)/(2)=(0.1)/(2)=0.05s` |
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| 226. |
Two simple harmonic motions act on a particle. These harmonic motions are `x=Acos (omegat+delta), y=Acos (omegat+alpha)` When `delta=alpha+(pi)/(20,` the resulting motion isA. an ellipse and the actual motion is counter clockwiseB. an ellipse and the actual motion is clockwiseC. a circle and the actual motion is counter clockwiseD. a circle and the actual motion is clockwise |
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Answer» Correct Answer - C `x=Acos(omegat+alpha+(pi)/(2))=-A sin (omegat+alpha)` `:. x^(2)+y^(2)=A^(2)[sin^(2)(omegat+alpha)+cos^(2)(omegat+alpha)]=A^(2)` which is an equation of a circle. Since y is `+ve` and `x` is `-ve` , so the circular motion is couter clockwise. |
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| 227. |
A particle is subjected to two simple harmonic motions in the same direction having equal amplitude and equal frequency. If the resultant amplitude is equal to the amplitude of individual motions, what is the phase difference between the motions. |
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Answer» Here, `a_(1)=r,a_(2)=r and R=r,theta=?` As` R^(2)=a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)costheta` `:. R^(2)=r^(2)+r^(2)+2r.rcostheta` `=2r^(2)(1+costheta)` or `1+costheta=(1)/(2)` or `costheta =(1)/(2)-1=-(1)/(2)=cos120^(@)` or `theta=120^(@)=2pi//3radian` |
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| 228. |
Two simple harmonic motions are represented by the following equations `y_1=10sin(3pit+pi/4)` `y_2=5(sin 3pit+sqrt3cos3pit)` Find out the ratio of their amplitudes. What are the time periods of two motions? |
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Answer» `y_(1)=10sin((pi)/(4))(12t+1)=10sin(3pit+(pi)/(4))` …(i) `y^(2)=5[sin3pit+sqrt(3)cos 3pit]` `=5xx2[sin3pitxx(1)/(2)+(sqrt(3))/(2)cos3pit]` `=10[sin3pit cos ((pi)/(3))+sin((pi)/(3))cos3pit]` `=10[sin(3pit+pi//3)]` …(2) The general equation of SHM is `y=Asin[omega t + phi_(0)]=A sin[(2pi)/(T)t+phi_(0)]` ...(iii) Comparing equation (i) and (ii) with equation (iii), we get `A_(1)=10, A_(2)=10, (2pi)/(T_(1))=3pi=(2pi)/(T_(2))` `:. (A_(1))/(A_(2))=(10)/(10)=1:1` and ` T_(1)=(2pi)/(3pi)=(2)/(3)s=T_(2)` |
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| 229. |
Two simple harmonic motions are represented by `y_(1)=4sin(4pit+pi//2)` and `y_(2)=3cos(4pit)`. The resultant amplitude isA. 7B. 1C. 5D. `(2+sqrt(3))` |
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Answer» Correct Answer - A `7` |
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| 230. |
A simple harmonic travelling wave is represented by `y=50sinpi(20t-0.08x)` is SI units. Find its amplitude, time period and wavelength. Also, calculate maximum particle velocity. |
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Answer» The given equation is `y=50sinpi(20t-0.08x)` `=50sin(20pit-(8pi)/(100)x)` Compare it with standard equation of wave motion `y=rsin((20pit)/(T)-(2pix)/(lambda))` `r=50m, (2pi)/(T)=20pi, T=(2)/(20)=0.1s` `(2pi)/(lambda)=(8pi)/(100), lambda=25m` Maximum particle velocity `upsilon_(max.)=romega =r(2pi)/(T)` `=50xx20pi` `=1000xx3.1416=3141.6m//s` |
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| 231. |
Give below are some functions of x and t to represesnt the displacement of an elastic wave. `(i) ``y=5cos (4x) sin (20t)` (ii) `y=4sin(5x-t//2)+3cos(5x-t//2)` (iii) `y=10cos[(252-250)pit]cos[(252+2500pit]` (iv)` y=100cos(100pit+0.5x)` State which of these represent (a) a travelling wave along-x direction (b) a stationary wave (c) beats (d) a travelling wave along `+x` direction Give reasons for your answers. |
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Answer» The four given equations are: (i) `y=5cos 4xsin20t` (ii) `y=4sin(5x-t//2)+3cos(5x-t//2)` (iii) `y=10cos(252-250)pit cos(252+250)pit` (iv) `y=100cos(100pit +0.5x)` From the basic concepts , we know (a) Eqn. (iv) represents a wave travelling along `-x` direction. (b) Eqn. (i) represents a stationary wave as terms containing x and t are independent. (c) Eqn. (iii) represents beats involving sum and difference of two frequencies 252 and 250. (d) Eqn. (ii) represents a wave travelling along `+x ` direction. |
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| 232. |
A body oscillates with SHM according to the equation `x(t)=5cos(2pit+pi//4)` where t is in second and x in metres. Calculate (a) displacement at t=0 (b) time period (c) initial velocity |
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Answer» `x(t)=5cos(2pit+pi//4)` …(i) (a) At `t=0,` displacement, `x(0)=5cos(2pixx0+pi//4)` or `x(0)=5cospi//4=(5)/(sqrt(2))m` (b) Comparing (i) with the standard equation of SHM. `x(t)=Acos(omegat+phi_(0)),` we have `omega=2pi or (2pi)/(T)=2pi or T=1s` (c) Velocity, `upsilon=(dx)/(dt)` `=[-5sin(2pit+pi//4)]xx2pi` At `t=0,` initial velocity, `upsilon_(0)=[-5sin(2pixx0+pi//4)]xx2pi` `=(-10pi)/(sqrt(2))=-5sqrt(2)pims^(-1)` |
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| 233. |
A wave in a string has an amplitude of `2 cm`. The wave travels in the `+ve` direction of x axis with a speed of 1`28 ms^-1` and it is noted that `5` complete waves fit in `4 m` length of the string. The equation describing the wave isA. `y=(0.02) m sin (15.7x-2010t)`B. `y=(0.02) m sin (15.7x+2010t)`C. `y=(0.02) m sin (7.85x-1005t)`D. `y=(0.02) m sin (7.85x+1005t)` |
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Answer» Correct Answer - C Here, amplitude `r=2cm =0.02m, upsilon=128m//s` `lambda=(4)/(5)=0.8m, v=(upsilon)/(lambda)=(128)/(0.08)=160Hz,` `omega=2piv=2xx(22)/(7)xx160=1005` and `k=(2pi)/(lambda)=(2xx(22//7))/(0.8)=7.85` Now `y=rsin (kx-omegat)` `=0.02 sin (7.85x-1005t)` |
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| 234. |
The equation of a wave on a string of linear mass density `0.04 kgm^(-1)` is given by `y = 0.02(m) sin[2pi((t)/(0.04(s)) -(x)/(0.50(m)))]`. Then tension in the string isA. `12.5N`B. `0.5N`C. `6.25N`D. `4.0N` |
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Answer» Correct Answer - C `y=0.02(m) sin[2pi((t)/(0.04(s))-(x)/(0.50(m)))]` Comparing it with the equation of wave `y=rsin ((2pi)/(lambda))(upsilont-x)=rsin 2pi((upsilont)/(lambda)-(x)/(lambda))` We have, `(upsilon)/(lambda)=(1)/(0.04)` and `lambda=0.50` `:. upsilon=(lambda)/(0.04)=(0.50)/(0.04)=(25)/(2)m//s` If `T` is tension in the string , then `upsilon=sqrt((T)/(m))` or`T= upsilon^(2)m=((25)/(2))^(2)xx0.04=6.25N` |
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| 235. |
The transverse displacement `y(x, t)` of a wave on a string is given by `y(x, t)= e ^(-(ax^(2) + bt^(2) + 2sqrt((ab))xt)`. This represents a :A. wave moving in `-x` direction with speed `sqrt((a)/(b))`B. standing wave of frequency `sqrt(b)`C. standing wave of frequency `(1)/(srt(b))`D. wave moving in `+x` direction with `sqrt((a)/(b))` |
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Answer» Correct Answer - A `y_((x,t))=e^(-(ac^(2)+bt^(2)+2sqrt(ab)xt))` `=e^(-(sqrt(a)x+sqrt(b)t)^(2))=e^(-(sqrt(b)t+sqrt(a)x)^(2))` It respresents a transverse wave, where velcoity of wave, `upsilon=(omega)/(K)=(sqrt(b))/(sqrt(a))` in `-ve x-` direction . Thus, options `(a)` is correct. |
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| 236. |
Awire having a linear mass density of `5.0xx10^(-3)kgm^(-1)` is stretched between two rigid supports with a tension of 45N. The wire resonates at a frequency of 420Hz. The next higher frequency at which the wire resonates is 490Hz. Find the length of the wire. |
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Answer» Here, `m=5.0xx10^(-3)kg//m.` Let l be the length of the wire vibrating in p loops, `T=450N` `:. 420=(p)/(2l)sqrt((T)/(m))` …(i) and `490=(p+1)/(2l)sqrt((T)/(m))` Dividing, we get `(420)/(490)=(p)/(p+1)=(6)/(7)` `7p=6p+6 or p=6.` From (i) , `420=(6)/(2l)sqrt((450)/(5.0xx10^(-3)))=(3)/(l)xx300` `l=(900)/(420)=2.14m` |
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| 237. |
A wire having a linear mass density of `5.0xx10^(-3)kgm^(-1)` is stretched between two rigid supports with a tension of 45N. The wire resonates at a frequency of 420Hz. The next higher frequency at which the wire resonates is 490Hz. Find the length of the wire. |
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Answer» Suppose the wire vibratees at a frequency of 420 Hz is nth harmonic and at a frequency of 490 Hz in `(n+10)` the harmonic. As, `v=(n)/(2l)sqrt((T)/(m)):. 420=(n)/(2l)sqrt((T)/(m))` …(i) `490=(n+1)sqrt((T)/(m))` `:. (490)/(420)=(n+1)/(n) :. n=6` From (i) , `420=(6)/(2l)sqrt((450)/(5xx10^(-3)))=(900)/(l)` `l=(900)/(420)=2.1m` |
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| 238. |
The length of a sonometer wire between two fixed ends is 110cm. Where should the two bridges the placed so as to divide the wire into three segments, whose fundamental frequencies are in the ration` 1:2:3?` |
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Answer» Here, total length of the wire, `L=110 cm, v_(1):v_(2):v_(3)=1:2:3` If `l_(1),l_(2),l_(3)` are the lengths of the three parts, then as `vprop(1)/(l)` `:. l_(1):l_(2):l_(3)=(1)/(1):(1)/(2):(1)/(3) =6:3:2` sum of the ratios`=6+3+2=11` `:. l_(1)=(L)/(11)xx6=(110)/(11)xx6=60cm ,` `:. l_(2)=(L)/(11)xx3=(110)/(11)xx3=30cm ,` `:. l_(2)=(L)/(11)xx2=(110)/(11)xx2=20cm,` Hence , the bridges should be placed at 60cm and 90 cm from the zero end of the wire. |
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| 239. |
A string vibrates in three loops when 8gram weight is placed in the pan. What mass should be kept in the pan to make it vibrate in six loops. Neglect the mass of string and mass of pan. |
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Answer» When string vibrates in p loops, its frequency `v=(p)/(2l)sqrt((T)/(m))` As v,l,T and m are constants, therefore, `psqrt(T)=` constant or `p^(2)T=` constant. `:. (T_(2))/(T_(1))=(p_(1)^(2))/(p_(2)^(2))=(3^(2))/(6^(2))=(1)/(4), T_(2)=(T_(1))/(4)=(8)/(4)=2g` |
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| 240. |
Equation for two waves is given as `y_(1)=asin(omegat+phi_(1)), y_(2)=asin(omegat+phi_(2))`. If ampitude and time period of resultant wave does not change, then calculate `(phi_(1)-phi_(2))`. |
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Answer» The equation of resultant wave is , `y=y_(1)+Y_(2)=asin(omegat+phi_(1))+asin(omegat+phi_(2))` `=2asin(((omegat+phi_(1))+(omegat+phi_(2)))/(2))` `cos(((omegat+phi_(1))-(omegat+phi_(2)))/(2))` `=2acos((phi_(1)-phi_(2))/(2))sin[omegat+(phi_(1)+phi_(2))/(2)]` The amplitude of the resultant wave is, `A=2acos((phi_(1)-pphi_(2))/(2))` ltbgt Given, `A=a, ` then `a=2acos((phi_(1)-phi_(2))/(2))` or `cos((phi_(1)-phi_(2))/(2))=(1)/(2)=cos60^(@)` or `(phi_(1)-phi_(2))/(2)=60^(@) or phi_(1)-phi_(2)=120^(@)` |
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| 241. |
Two particles P and Q are executing SHM across same straight line whose equations are given as `y_(P)=Asin (omegat+phi_(1)) and y_(Q)=Acos (omegat+phi_(2))` An observer, at `t=0` observer the particle P at a distance `A//sqrt(2)` moving to the right from mean position O while particle Q at `(sqrt(3))/(2)` A moving to the left from mean position O as shown in figure. Then `phi_(2)-phi_(1)(i n rad)` ie equal to A. `5pi//6`B. `3pi//4`C. `5pi//2`D. `7pi//12` |
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Answer» Correct Answer - D `y_(p)=a sin (omegat+phi_(1)),` For particle P, `t=0, y_(p)=(A)/(sqrt(2))` `:. (A)/(sqrt(2))=A sin (0+phi_(1))=Asin phi_(1)` or `sin phi_(1)=(1)/(sqrt(2)) or phi_(1)=(pi)/(4)rad` `y_(Q)=A cos (omegat +phi_(2)),` For particle `Q,t=0, y_(Q)=(-sqrt(3))/(2)A` `:. -(sqrt(3))/(2)A=A cos (0 +phi_(2))=A cosphi_(2)` or `cosphi_(2)=-(sqrt(3))/(2)=cos((5)/(6))pi or phi_(2)=(5pi)/(6)` `:. phi_(2)-phi_(1)=(5pi)/(6)-(pi)/(4)=(7pi)/(12)rad` |
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| 242. |
At concerts, the audience in balcony, while jumping up and down in tune with the music may cause damage to balcony, sometimes. Why? |
| Answer» When audience sitting in balcony jump up and down in tune with the music, sometimes, the jumping frequency may match the natural vibration frequency of the balcony. Due to it, resonance takes place. The balcony starts vibrating with large amplitude and might be damaged. | |
| 243. |
Two pipes have equal length of 2m, one is closed at one end and the other is open at both ends. The speed of sound in air is `340m//s`. At what frequency will both the pipes resonate ? |
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Answer» Here, `l=2m, upsilon=340m//s.` Resonant frequencies for closed organ pipe are `f_(1)=((2n_(1)-1)upsilon)/(4l), ` where `n_(1)=1,2,3…` `f_(1)=(upsilon)/(4l), (3upsilon)/(4l),(5upsilon)/(4l)` …etc. Resonant frequencies for open organ pipe are `f_(2)=(n_(2)upsilon)/(2l),` where `n_(2)=1,2,3.....` `:. f_(2)=(upsilon)/(2l),(2upsilon)/(2l), (3upsilon)/(2l), (4upsilon)/(2l)` ...etc. Comparing values of `f_(1)` and `f_(2)`, we find that there is no common frequency at which the two pipes can resonate. |
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| 244. |
A point performs damped oscillations with frequency `omega` and damping coefficient `beta` according to the (4.1b). Find the initial amplitude `a_(0)` and the initial phase `alpha` if at the moment `t=0` the displacement of the point and its velocity projection are equal to (a) `x(0)=0` and `u_(x)(0)=dot(x_(0))` , (b) `x(0)=x_(0)` and `v_(x)(0)=0.` |
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Answer» We write `x=a_(0)e^(-betat) cos ( omega t+alpha)`. `x(0) =0implies alpha=+ (pi)/(2)implies x =+- omega a_(0)` Since `a_(0)` is `+ve` , we must choose the upper sign if `dot(x) (0)lt 0` and the lower sign if `dot(x) (0)gt 0` . Thus `a_(0)=(|dot(x) (0)|)/(omega)` and `{(+(pi)/(2) if dot (x) (0) lt 0),(-(pi)/(2)if dot(x)(0) gt 0):}` `(b)` we write `x=Re A e^(-beta t + i omegat), A=a_(0)e^(ialpha)` Then `dot (x) =v_(x)=Re (-beta+iomega) A e^(- beta t +iomegat)` From `v_(x)(0)=0` we get `Re ( - beta + i omega)A=0` This implies `A=+- i(beta+iomega)` where `B` is real and positive. Also `x_(0) R eA=+_ omega B` Thus `B=(|x_(0)|)/(omega )` with `+` sing in `A` if `x_(0) lt 0` `- sin` in `A` if `x_(0) gt 0` So `A=+-I (beta+iomega)/(omega)|x_(0)|=(+-1++-(ibeta)/(omega))|x_(0)|` Finally `a_(0)=sqrt(1+((beta)/(omega))^(2))|x_(0)|` `tan alpha=(-beta)/(omega), alpha=tan ^(-1)((-beta)/(omega))` `alpha` is in the `4^(th)` quadrant `(-(pi)/(2) lt alpha lt (0))` if `x_(0) gt 0` and `alpha` is in the `2 ^(nd)` quadrant `((pi)/(2),alpha lt pi)` if `x_(0) lt 0` |
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| 245. |
A block whose mass is 1 kg is fastened to a spring.The spring has a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface at `t=0`. Calculate the kinetic, potential and total energies of the blocak when it is 5cm away from the mean position. |
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Answer» Since, time is noted from the equilibrium position for SHM, hence displacement, x at time t is `x=x_(0)sinomegat,` where ` omega=sqrt((k)/(m))=sqrt((50)/(1))Nm^(-1)` Velocity, `V=(dx)/(dt)=x_(0)omega cosomegat` When `x=5cm=0.50m` and `x_(0)=10cm=0.10m` Then, `0.05=0.10sinomegat` or `sinomegat=(1)/(2)=sin((pi)/(6))` or `omega=(pi)/(6)` `:. cosomegat=cos((pi)/(6))=(sqrt(3))/(2)` Hence, `V=0.1XXsqrt(50)xxcos((pi)/(6))` `=0.1xxsqrt(50)xx(sqrt(3))/(2)=0.6123m//s` K.E. of block, `E_(K)=(1)/(2)mV^(2)` `=(91)/(2)xx1xx(0.6123)^(2)=0.1875J` P.E. of block , `E_(P)=(1)/(2)kx^(2)` `=(1)/(2)xx50xx(0.05)^(2)=0.0625J` Total energey `=E_(K)+E_(P)=0.875+0.0625` `=0.25J` |
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| 246. |
A block of mass one kg is fastened to a spring with a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface from rest at `t=0. ` Write the expression for its x(t) and v(t). |
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Answer» Here, `m=1kg,k-50Nm^(-1)` `a=10cm=0.10m` `omega=sqrt((k)/(m))=sqrt((50)/(1))=7.07rad//s` Since the motion starts from th emean position, so the displacement equation can be given as `x(t)=asinomegat=-.10sin7.07tm` and velocity, `v(t)=dx//dt=aomega cosomegat` `=0.10xx7.07 cos 7.07t ` `=0.707cos7.07tms^(-1)` |
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| 247. |
The equation of a travelling plane sound wave has the form `y=60 cos (1800t-5.3x)`, where y is in micrometres, t in seconds and x in metres. Find (a). The ratio of the displacement amplitude with which the particle of the medium oscillate to the wavelength, (b).the velocity oscillation ammplitude of particles of the medium and its ratio to the wave propagation velocity ,(c). the particle acceleration amplitude. |
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Answer» The given wave equation `xi=60 cos (1800t-5.3x)` is of the type `xi= a cos ( omegat-kx), `where` a=60xx10^(-6)m` `omega=1800 ` per sec and `k=5.3 `per metre As `k=(2pi)/(lambda), `so` lambda=(2pi)/( k)` and also `k=(omega)/(v), `so` v=(omega)/(k)=340m//s` `(a)` Sought ratio `=(a)/(lambda)=(ak)/(2pi)=5.1xx10^(-5)` `(b)` Since `xi=a cos ( omegat-kx)` `(deltaxi)/(deltat)=-a omegasin (omegat-kx)` So velocity oscillation amplitude `(deltaxi)/(deltat)_(m)` or `v_(m)=a omega=0.11m//s ...(1)` and sought ratio of velocity oscillation amplitude to the wave propagation velocity `((v_(m))/(v))=(0.11)/(340)=3.2xx10^(-4)` `(c)` Relative deformation `=( delta_(m))/(v)=a k sini (omegat-kx)` So, relative deformation amplitude `=((delta xi)/(deltax))=ak=(60xx10^(-6)xx5.3)m=3.2xx10^(-4)m ...(2)` From Eqns `(1)` and `(2)` `((deltaxi)/(deltax))_(x)=ak=(aomega)/(v)=(1)/(v)((deltaxi)/(deltat))_(m)` Thus` ((deltaxi)/( deltax))_(m)=(1)/(v)((deltaxi)/(deltat)), `where` v= 340 m//s` is the wave velocity. |
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| 248. |
A girl is sitting on a swing. Another girl sits by her side. What will be the effect on the periodic-time of the swing? |
| Answer» There will be no change in the periodic time because the periodic time is independent of mass but depends upon length of pendulum and acceleration due to gravity at a place, which are not affected by sittin another girl on a swing. | |
| 249. |
A silver atom in a solid oscillates in S.H.M. in a certain direction with a frequency of `10^(12)s^(-1)`. What is the force constant of the bonds connecting one atom with the other ? Molecular weight of silver `=108` and Avogadaro number `=6.2xx10^(23) (gm mo l e)^(-1)`. |
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Answer» Mass of one silver atom, `m=(M)/(N)=(108)/(6.02xx10^(23))=1.794xx10^(-22)g=1.794xx10^(-25)kg` Frequency of S.H.M. is , `v=(1)/(2pi)sqrt((K)/(m))` `:.` Force constant, `K=4pi^(2)mv^(2)=4xx(3.14)^(2)xx(1.794xx10^(-25))(10^(12))^(2)` `=7.1Nm^(-1)` |
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| 250. |
A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing. (a) the frequency of oscillation of the mass. (b) the maximum acceleration of the mass. (c) the maximum speed of the mass. |
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Answer» Here, `k=1200Nm^(-1),m=3.0kg, a=2.0cm=0.2m` (a) Frequency, `v=(1)/(T)=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx3.14)sqrt((1200)/(3))=3.2s^(-1)` (b) Acceleration, `A=omega^(2)y=(k)/(m)y` Acceleration will be maximum when y is maximum, i.e., `y=a` `:. ` Max. acceleration, `A_(max)=(ka)/(m)=(1200xx0.02)/(3)=8ms^(-2)` (c) Max. speed ofo the mass will be when it is passing through the mean position, which is given by `V_(max)=aomega=asqrt((k)/(m))=0.02xxsqrt((1200)/(3))=0.4ms^(-1)` |
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