MCQ Questions for Class 11 Engineering Physics Waves Quiz 10

$x_1 = A\sin(\omega t - 0.1x)$ and

$x_2 = A\sin(\omega t - 0.1x - \dfrac{\phi}{2})$ . The resultant amplitude of combined wave is:-

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$2 A\cos \dfrac{\phi}{4}$
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$A \sqrt{2\cos \dfrac{\phi}{2}}$
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$2 A\cos \dfrac{\phi}{2}$
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$A \sqrt{2(1 + \cos \dfrac{\phi}{4})}$

Explanation

$x_1 = A sin (\omega t- 0.1 x)$

$x_2 = a sin(\omega t-01x-\phi /2)$

$x_1+x_2 = A(sin(\omega t-0.1x)+sin(\omega t-0.1x-\phi /2))$

$= A (2sin \dfrac{(\omega t-0.1x)+(\omega t-0.1x-\phi /2)}{2}$

$cos \dfrac{(\omega t-0.1x)-(wt-01x-\phi /2)}{2})$

$= (2A cos \phi /4) (sin(\omega t-0.1x - \phi /2))$

Amplitude =

$2A cos \phi /4$

1186630_1139896_ans_c662794dcab24203a6098e9e2290f301.jpg

Two parallel beams of light of wavelength

$\lambda$ , inclined to each other at angle

$\theta \ ( < + 1)$ are inclined on a plane at near normal incidence. The fringe width will be.

0%
$\dfrac{\lambda }{{2\theta }}$
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$\dfrac{2\lambda }{{\theta }}$
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$\dfrac{\lambda }{{\theta }}$
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$2 \lambda \ \sin \theta$

Two waves

$E _ { 1 } = E _ { 0 } \sin \omega t$ and

$E _ { 2 } = E _ { 0 } \sin ( \omega t + 60 )$ superimpose each other. Find out initial phase of resultant wave?

0%
$30 ^ { \circ }$
0%
$60 ^ { \circ }$
0%
$120 ^ { \circ }$
0%
$0 ^ { \circ }$

Explanation

Let the resultant wave be

$E=E0' \sin (wt+\phi)$ ./ Then

$E=E_1+E_2$

$E=E_0\sin wt +E_0\sin (wt+60^o)$

$E=E_0(\sin wt +\sin (wt +60^o)$

$\sin \cos +\sin (B)=2\sin \left (\dfrac {A+B}{2}\right) \cos \left (\dfrac {B-A}{2}\right)$

$\sin wt +\sin (wt +60^o)=2 \sin (wt +30^o).\cos (wt)$

So,

$E=2E_0 \cos (wt) \sin (wt+30^o)$

So,

$E_0=2E_0 \cos wt$ and

$\phi =30^o$

Option

$A$ is correct

The amplitude of a particle due to superposition of following S.H.Ms. Along the same line is

${ X }_{ 1 }=2sin50\pi t\quad ;\quad { X }_{ 2 }=10sin\left( 50\pi t+{ 37 }^{ } \right)$

${ X }_{ 3 }=-4sin50\pi t\quad ;\quad { X }_{ 4 }=-12sin50\pi t$

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$4\sqrt { 2 }$
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$4$
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$6\sqrt { 2 }$
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None of these

Equation of two S.H.M.

$x_1 = 5 sin ( 2 \pi t + \pi /4),$

$x_2 = 5\sqrt { 2 } (sin\quad 2\pi t+cos2\pi t)$ ratio of amplitude & phase difference will be

0%
$2 : 1, 0$
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$1 : 2, 0$
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$1 : 2, \pi /2$
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$2 : 1, \pi / 2$

Explanation

We have,

$X_{1}=5\sin(2nt+n/4)$

$A_{1}=5\ m$ and

$\phi_{1}=n/4$

$X_{2}=5\sqrt{2}(\sin 2nt+\cos 2nt)$

$X_{2}=5\sqrt{2}\times \sqrt{2}\left(\dfrac{\sin 2nt}{\sqrt{2}}+\dfrac{\cos 2nt}{\sqrt{2}}\right)$

$X_{2}=10\sin(2nt+\pi/4)$

$A_{2}=10\ m$ and

$\phi_{2}=\pi/4$

$\dfrac{A_{1}}{A_{2}}=\dfrac{5 m}{10 m}=\dfrac{1}{2}$ and

$\Delta\phi=\phi_{1}-\phi_{2}=0$

Option

$B$ is correct.

Two simple harmonic motions are represented by the equation

$y_{1} = 10\sin (4\pi t + \pi/4)$ and

$y_{2} = 5(\sin 3\pi t + \sqrt {3} \cos 3\pi t)$ . Their amplitudes are in the ratio.

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$1 : 1$
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$2 : 1$
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$2 : \sqrt {3}$
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$\sqrt {3} : 2$

A progressive wave moves with a velocity of

$36 \mathrm { m } / \mathrm { s }$ in a medium with a frequency of 200Hz. The phase difference betveen two particles seperated by a distance of

$1$

$cm$ is

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$40 ^ { \circ }$
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$20 \mathrm { rad }$
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$\dfrac { \pi } { 9 } \mathrm { rad }$
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$\dfrac { \pi } { 9 } ^{o}$

A simple harmonic oscillation is represented by the equation y

$= 0.4 sin \left ( \frac{440t}{7}+0.61x \right )$ . Where 'Y' and 'X' are in m and time 't' in seconds respectively. The value of time period in seconds

0%
0.1
0%
0.01
0%
1
0%
10

Explanation

It is given that,

$y=0.4\sin \left( \dfrac{440}{7}t+0.61x \right)$

From the above equation:

$\omega =\dfrac{440}{7}$

$\dfrac{2\pi }{T}=\dfrac{440}{7}$

$\dfrac{2\times 22}{7T}=\dfrac{440}{7}$

$T=0.1\,s$

If a body of mass 36gm moves with S.H.M of amplitude A =13 cm and period T= 12 sec. At a time t= ) the displacement is x = +13 cm.The shortest time of passage from x = +6.5 cm to x = -6.5 is:

0%
4 sec
0%
2 sec
0%
6 sec
0%
3 sec

Explanation

The equation of the SHM is:-

$y=A\cos \omega t$ where

$A=13cm$ and

$\dfrac{2\pi}{T}$

$T=125 , \Rightarrow \omega=\dfrac{2\pi }{12}=\dfrac{\pi}{6}rad/s$

also at

$t=0,y=13$ as given

$=y=13 \cos \left(\dfrac{\pi t}{6}\right)$

$\Rightarrow t=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{y}{13}\right)$

when particle is at

$y=6.5$ ,time is,

$t_1=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{6.5}{13}\right)=\dfrac{6}{\pi}\times \dfrac{\pi}{3}=25$

also when

$y=-6.5$ , time is :-

$t_2=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{-6.5}{13}\right)=\dfrac{6}{\pi}\times \dfrac{2\pi}{3}=45$

$\therefore$ Time difference

$\Delta t=t_2-t_1=(4-2)=2s$

1082674_1172613_ans_93cdf61d582948138bc54f8df9bebc9c.png

Three waveforms travelling along a straight line have the forms:

$2A\sin \left (kx - \omega t + \dfrac {\pi}{3}\right ), \sqrt {3}A \cos \left (kx - \omega t - \dfrac {\pi}{3}\right ), 2\sqrt {3} \cos \left (kx - \omega t + \dfrac {\pi}{3}\right )$ . The amplitude of the resulting waveform is :

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$(2 + 3\sqrt {3})A$
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$\sqrt {31}A$
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$\sqrt {19}A$
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$(2 - \sqrt {3})A$

Two waves having equation

$x _ { 1 } = a \sin \left( \omega t + \phi _ { 1 } \right)$ and

$x _ { 2 } = a \sin \left( \omega t + \phi _ { 2 } \right)$ If in the resultant wave the frequency and amplitude remains equals to amplitude of superimposing waves. Then phase difference between them-

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$\dfrac { \pi } { 6 }$
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$\dfrac { 2 \pi } { 3 }$
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$\dfrac { \pi } { 4 }$
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$\dfrac { \pi } { 8 }$

Explanation

$x_{1}=a\, sin (\omega t+\phi _{1})$

$x_{2}= a\, sin (\omega t+\phi _{2})$

$\triangle \phi$ = phase difference =

$\phi _{1}-\phi _{2}$

Resulatant amplitude

$\Rightarrow a=\sqrt{(a)^{2}+(a)^{2}+2a^{2}cos \triangle \phi }$

$\Rightarrow a^{2}=2a^{2}+2a^{2}cos \triangle \phi$

$cos \triangle \phi = -\frac{1}{2}$

$\triangle \phi = \frac{2\pi}{3}$

Option B

1164570_1182159_ans_0d596e5911104bcd81119803014e32fc.jpg

Three simple harmonic motions in the same direction having the same amplitude A and same period are superposed. If each differs in phase from the next by

$45^0$ , then

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the resulting amplitude is $\Big ( 1 + \sqrt{3} \Big ) A$ .
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the resulting motion is not simple harmonic
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The energy associated with the resulting motion ( $3 + 2 \sqrt{2}$ ) times the energy associated with any single motion.
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The phase of the resultant motion relative to the first is $90^0$

Explanation

[C]

${Y_1} = A\sin \left( {wt - \dfrac{\pi }{4}} \right),\,\,{Y_2} = A\sin \left( {wt} \right),\,\,{Y_3} = A\left( {\sin wt + \dfrac{\pi }{4}} \right)$

on superimposing, Resultant SHM =

$\begin{array}{l} Y=A\left[ { 2\sin wt\cos \dfrac { \pi }{ 4 } +\sin wt } \right] \Rightarrow A\left[ { \sqrt { 2 } \sin wt+\sin wt } \right] \\ Y=A\left( { 1+\sqrt { 2 } } \right) \sin wt,\, \, { { Resultant } }\, \, AMP=\left( { 1+\sqrt { 2 } } \right) A \\ \frac { { \in { { resultant } } } }{ { \in { { single } } } } =\dfrac { { { { \left\{ { A\left( { 1+\sqrt { 2 } } \right) } \right\} }^{ 2 } } } }{ { { A^{ 2 } } } } \Rightarrow \in { { Resultant } }\, \, =\left( { 3+2\sqrt { 2 } } \right) { { single } } \end{array}$

Two particles are executing simple harmonic motion of the same amplitude

$A$ and frequency

$\omega$ along the

$x-axis.$ Their mean position is separated by distance

$X_0(X_0 > A)$ . If the maximum separation between them is

$(X_0 + A ),$ the phase difference between their motion is :-

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$\dfrac{\pi}{4}$
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$\dfrac{\pi}{6}$
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$\dfrac{\pi}{2}$
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$\dfrac{\pi}{3}$

Explanation

${X_1} = A\sin \left( {wt + {\phi _1}} \right)$

${X_2} = A\sin \left( {wt + {\phi _2}} \right)$

${X_1} - {X_2} = A\left[ {2\sin \left[ {wt + \frac{{{\phi _1}}}{{{\phi _2}}}} \right]\sin \left[ {\frac{{{\phi _1} - {\phi _2}}}{2}} \right]} \right]$

$A = 2A\sin \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right)$

$\frac{{{\phi _1} - {\phi _2}}}{2} = \frac{\pi }{6}$

${\phi _1} = \frac{\pi }{3}$

Hence,

option

$(D)$ is correct answer.

Two

$SHMs$ are given by

$Y_{1}= a\left[ \sin { \left( \dfrac { \pi }{ 2 } \right) } t+\varphi \right]$ and

$Y_{2}= b\sin { \left[ \left( \dfrac { 2\pi t }{ 3 } \right) +\varphi \right] }$ . The phase difference between these two after

$'1'\ sec$ is:

0%
$\pi$
0%
$\dfrac {\pi}{2}$
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$\dfrac {\pi}{4}$
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$\dfrac {\pi}{6}$

Equation of two

$S.H.M,\,\,{x_1} = 5\sin \left( {2\pi t + \pi /4} \right),\,{x_2} = 5$

$\sqrt 2 \left( {\sin 2\pi t + \cos 2\pi t} \right)$ . Ratio of amplitude & phase difference will be :

0%
$2:\,1,\,0$
0%
$1:\,2,\,0$
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$1:2,\,\pi /2$
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$2:1,\,\pi /2$

Explanation

Equation of S.H.M ,

$x_1=5sin(2\pi t +\pi/4)$ . . . . . . .(1)

$x_2=5\sqrt{2}(sin2\pi t+cos2\pi t)$

$x_2=10(\dfrac{1}{\sqrt{2}}.sin2\pi t+\dfrac{1}{\sqrt{2}}.cos2\pi t)$

$x_2=10(cos\dfrac{\pi}{4}.sin2\pi t+sin\dfrac{\pi}{4}.cos2\pi t)$

$x_2=10(sin2\pi t+\pi/4)$ . . . . . . . . ..(2)

The ratio of amplitude,

$a_1:a_2=1:2$

The phase difference is,

$\phi=\theta _1-\theta_2$

$\phi=\dfrac{\pi}{4}-\dfrac{\pi}{4}=0$

The correct option is B.

The distance between two consecutive crests in a wave train produced in string is 5 m. If two complete waves pass through any point per second, the velocity of wave is:

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$2.5 \mathrm { m } / \mathrm { s }$
0%
$5 \mathrm { m } / \mathrm { s }$
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$10 \mathrm { m } / \mathrm { s }$
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$15 \mathrm { m } / \mathrm { s }$

Explanation

Hint:- Two wavelength passes through a point in given time so apply formula of velocity of wave accordingly.

Step 1: Note the Given values

Two consecutive crest = $5\,m$ .

Two wave cross in time, $t=\,1\sec$

Wavelength = distance between two consecutive crest

$\lambda =5\,m$

Step 2: Calculate velocity of wave

\Velocity, $v=\dfrac{2\lambda }{t}=\dfrac{2\times 5}{1}=10\,m{{s}^{-1}}$

Hence, wave speed is $10\,m{{s}^{-1}}$

A body is on a rough horizontal surface which is moving horizontally in

$SHM$ of frequency

$2\ Hz$ . The coefficient of static friction between the body and the surface is

$0.5$ . The maximum value of the amplitude for which the body will not slip along the surface is approximately

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$9\ m$
0%
$6\ m$
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$4.5\ m$
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$3\ m$

A particle performing

$S.H.M$ is found at its equilibrium at

$t= 1\ s$ and it is found to have a speed of

$0.25\ m/s$ at

$t=2\ s$ .If the period of oscillation is

$6\ s$ .Calculate amplitude of oscillation

0%
$\dfrac { 3 }{ 2\pi } m$
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$\dfrac { 3 }{ 4\pi } m$
0%
$\dfrac { 6 }{ \pi } m$
0%
$\dfrac { 3 }{ 8\pi } m$

Explanation

Given,

$\omega=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$

Let the displacement of particle be,

$y=A\sin(\omega t+\phi)$

$t=1,y=0$

$\dfrac{\pi t}{3}+\phi=\pi$

$\phi=\dfrac{2\pi}{3}$

Velocity at

$t=1, 0.25m/s$

$v=A\omega\cos\left(\dfrac{2\pi}{3}+\dfrac{2\pi}{3}\right)=\dfrac{A\omega}{2}$

$0.25=\dfrac{\pi A}{6}$

$A=\dfrac{3}{2\pi}$

Option

$\textbf A$ is the correct answer

$4$ kg particle is moving along the x-axis under the action of the force

$F=-\left(\dfrac{\pi^2}{16}\right)\times N$ . At

$t=2$ sec, the particle passes through the origin and at

$t=10$ sec its speed is

$4\sqrt{2}$ m/s. The amplitude of the motion is?

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$\dfrac{32\sqrt{2}}{\pi}$ m
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$\dfrac{16}{\pi}$ m
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$\dfrac{4}{\pi}$ m
0%
$\dfrac{16\sqrt{2}}{\pi}$ m

The equation of a progressive wave are

$Y=sin[200n(t-\cfrac x {330})]$ ,where x is in meter and f is in second. The frequency and velocity of wave are

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$100 Hz,5m/s$
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$300 Hz,100m/s$
0%
$100 Hz,330 m/s$
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$30 m/s,5Hz$

An object of mass m is attached to a spring.The restoring force of the spring is F=

$- \lambda {x^3},$ where x is the displacement. the oscillation period depends on the mass,

$\lambda$ and oscillation amplitude.suppose the object is initially at rest.If the initial displacement is D then its period is

$\tau$ .If the initial displacement is 2D, find the period.(Hint: Use dimension analysis.)

0%
8 $\tau$
0%
2 $\tau$
0%
$\tau$
0%
$\tau$ /2

Three waves of amplitude

$10\mu\ m,4\mu\ m$ and

$7\mu\ m$ arrive at a point with successive phase difference of

$\pi/2$ . The amplitude of the resultant wave is

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$2\mu m$
0%
$7\mu m$
0%
$5\mu m$
0%
$1$

Explanation

Given,

Amplitude of the waves are,

$a_1=10\mu m,\,a_2=4\mu m,\,a_3=7\mu m$

and the phase difference between 1st and 2nd wave is

$\dfrac{\pi}{2}$ and that between 2nd and 3rd wave is

$\dfrac{\pi}{2}$ . Therefore, phase difference between 1st and 3rd is

$\pi$ .

Combining 1st with 3rd, their resultant amplitude will be

$A_1^2=a_1^2+a_3^2+2a_1a_3cos\phi$

$A_1=\sqrt{10^2+7^2+2\times 10\times 7cos\pi}$

$=\sqrt{100+49-140}=\sqrt 9 =3\mu m$ in the direction of first.

Now combining this with 2nd wave we get the resultant wave,

$A^2=A_1^2+a_2^2+2A_1a_2 \dfrac{cos\pi}{2}$

$A=\sqrt{3^2+4^2+2\times 3\times 4cos90^2}=\sqrt{9+16}=5\,\mu m$

A transverse progressive wave on a stretched string has a velocity of

$10ms^{-1}$ and frequency of

$100Hz$ . The phase difference between two particles of the string which nbare

$2.5cm$ apart will be :

0%
$\cfrac{\pi}{8}$
0%
$\cfrac{\pi}{4}$
0%
$\cfrac{3\pi}{8}$
0%
$\cfrac{\pi}{2}$

Explanation

Given,

$v=10m/s$ velocity

$f=100Hz$ frequency

$\Delta=2.5cm$ path difference

wavelength,

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{10}{100}=0.1m$

Phase difference,

$\phi=\dfrac{2\pi}{\lambda}\times \Delta$

$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$

$\phi=\dfrac{\pi}{2}$

The correct option is D.

In a string the speed of wave is

$10m/s$ and its frequency is

$100$ Hz. The value of the phase difference at a distance

$2.5$ cm will be :

0%
$\pi/2$
0%
$\pi/8$
0%
$3\pi/2$
0%
$4\pi$

Explanation

Given,

$f=100Hz$

$x=2.5cm$

$v=10m/s$ path difference

wavelength,

$\lambda =\dfrac{v}{f}=\dfrac{10}{100}=0.1m$

The phase difference,

$\phi=\dfrac{2\pi}{\lambda}\times x$

$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$

$\phi=\dfrac{\pi}{2}$

The correct option is A.

A wave is travelling along a string. At an instant shape of the string is as shown in figure. At this instant, point A is moving upwards. Which of the following statements is/are correct ?

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The wave is travelling to the right
0%
Displacement amplitude of the wave is equal to displacement of B at this instant
0%
At this instant velocity of C is also directed upwards
0%
Phase difference between A and C may be equal to $\pi/2$

Explanation

$\textbf{Explanation}$

1) The velocity of

$A$ is upward only if Wave moves to the left

2) Displacement of

$B$ from mean position is the Amplitude of wave.

3) as Wave is propogating towards left,

$v_{c}$ is downwards.

4) We don't know distance between

$A, C$

$\triangle \phi = \dfrac {2\pi}{\lambda} (\triangle x)$

$\therefore \triangle \phi$ cannot be known

Answer B.

2162696_1215906_ans_0c48dd523e304c9aa00dc24bfb5ca4f9.jpg

Two waves having intensity ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is:

0%
10 : 8
0%
9 : 1
0%
4 : 1
0%
2 : 1

Explanation

Hint:

If two waves of intensity ${{I}_{1}}\,and\,{{I}_{2}}$ interfere, then the ration of maximum to minimum intensity in interference is given as,

$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}$

Step 1: Assume values of intensities.

Given,

Ratio of Intensities of two waves ${{I}_{1}}\,and\,{{I}_{2}}$

${{I}_{1}}:{{I}_{2}}=9:1$ .

$\Rightarrow {{I}_{1}}=9x,\,{{I}_{2}}=1x$

for any constant $x$ .

Step 2: Calculate ratio of intensities.

The ratio of maximum to minimum intensity in interference is given as,

$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}$

$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{9x}+\sqrt{1x}}{\sqrt{9x}-\sqrt{1x}} \right)}^{2}}$

$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{4}{2} \right)}^{2}}$

$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{4}{1}$$

$\Rightarrow {{I}_{\max }}:{{I}_{\min }}=4:1$

The required ratio is 4:1. Option C.

A particle is describing SHM with amplitude

$'a'.$ When the potential energy of particle is one fourth of the maximum energy during oscillation, then its displacement from mean position will be:

0%
$\dfrac {a}{4}$
0%
$\dfrac {a}{3}$
0%
$\dfrac {a}{2}$
0%
$\dfrac {2a}{3}$

The irreducible phase difference in any wave of 5000 A from a source of light is

0%
$\pi$
0%
$12\pi$
0%
$12\pi \times{10}^{6}$
0%
$\pi \times {10}^{6}$

A stretched sting of length L,fixed at both ends can sustain stationary waves of wavelength

$\lambda$ Choose which of the following value of wavelength is not possible?

0%
2 L
0%
4 L
0%
L
0%
$\dfrac { L }{ 2 }$

The amplitude of a

$SHM$ reduces to

$1/3$ in first

$20$ second then in first

$40$ second its amplitude becomes:

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{9}$
0%
$\dfrac{1}{27}$
0%
$\dfrac{1}{{\sqrt 3 }}$

The ratio longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

0%
$\dfrac{4}{3}$
0%
$\dfrac{525}{376}$
0%
$25$
0%
$\dfrac{900}{11}$

Explanation

$\textbf{Hint:}$ Use hydrogen atom spectrum wavelength formula.

$\textbf{Step 1:}$ The wave number is given by

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

For shortest wavelength,

$n_1=infinite\, n_2=1$

$\dfrac{1}{\lambda _{min}}=R(\dfrac{1}{1^2}-\dfrac{1}{(infinite)^2})=R$

For longest wavelength,

$n_1=6$ ,

$n_2=5$

$\dfrac{1}{\lambda _{max}}=R(\dfrac{1}{5^2}-\dfrac{1}{6^2})=\dfrac{11R}{900}$

$\textbf{Step 2:}$ Ratio,

$\dfrac{1/\lambda _{min}}{1/\lambda_{max}}=\dfrac{R}{11R/900}$

$\dfrac{\lambda _{max}}{\lambda_{min}}=\dfrac{900}{11}$

$\textbf{Option D is correct.}$ .

Three waves of amplitude

$10\mu m,4\mu m$ and

$7\mu m$ arrival at a point with successive phase difference of

$\pi /2$ . The amplitude of the resultant wave is

0%
$2\mu m$
0%
$7\mu m$
0%
$5\mu m$
0%
$1$

Equations

$Y_1=a$ sin

$\omega t$ and

$Y_2=a/2$ sin

$\omega t + a/2$ cos wt represent S.H.M. The ratio of the amplitude of first to that of second S.H.M. is

0%
1
0%
0.5
0%
2
0%
$\sqrt 2$

the diagram shows the propagation of a wave. Which points are in same phase ?

0%
F and G
0%
C and E
0%
B and G
0%
B and F

Explanation

Hint: Points moving with same velocity are in phase.

Correct answer: Option D

Explanation:

Points B and F are in same phase as they are $\lambda$ distance apart.

$\lambda$ is the Wavelength, distance between corresponding points of two consecutive waves.

Corresponding points are those points that have completed identical fractions of their periodic motion.

Two waves are represented as

$y_1 = 2a sin (\omega t + \pi/6)$ and

$y_2 = -2a cos \Big \lgroup \omega t - \frac{\pi}{6} \Big \rgroup$ . The phase difference between the two waves is

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{4\pi}{3}$
0%
$\dfrac{3\pi}{3}$
0%
$\dfrac{5\pi}{6}$

The time lag between two particles vibrating in a progressive wave separated by a distance

$20$ m is

$0.02 s$ . The wave velocity if the frequency of the wave is

$500 Hz$ , is

0%
$1000 ms^{-1}$
0%
$500 ms^{-1}$
0%
$2000 ms^{-1}$
0%
$250 ms^{-1}$

Explanation

Given,

distance,

$s=20m$

$t=0.02s$

$f=500Hz$

Velocity,

$v=\dfrac{s}{t}$

$v=\dfrac{20}{0.02}=1000m/s$

The correct option is A.

A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest, it travels a distance

$a$ and in the next second, it travels a distance

$b$ in the same direction. The amplitude of the motion is:

0%
$\cfrac {2a^2}{3b-a}$
0%
$\cfrac {3a^2}{3a-b}$
0%
$\cfrac {2a^2}{3a-b}$
0%
$\cfrac {3a^2}{3b-a}$

Explanation

Let the acceleration be

$f, f=\omega^2 x$
Therefore, distance of the particle from the centre at any time

$t$ is given by

$x=r \cos (\omega t)$ , where

$r$ is the amplitude
when

$t=1\ s, x=r-a$

$\therefore \ (r-a)=r\cos \omega$

$\cos \omega =\dfrac {r-a}{r}$
When

$t=2\ s, x=r-a-b$ ,
therefore

$r-a-b=\cos 2\omega$

$\therefore \ r-a-b=r(2\cos^2 \omega -1)$
Substituting the value of

$\cos \omega$ from equation.
we get

$r-a-b=r \left[2\dfrac {(r-a)^2}{r^2}-1\right]$

$r-a-b=\dfrac {2(r-a)^2}{r}-r$

$\therefore \ r(3a-b)=2a^2 \ \Rightarrow \ r=\dfrac {2a^2}{3a-b}$

A particle executes SHM of amplitude A if

$T_1$ and

$T_2$ are the time taken by the particle to traverse from 0 to A/2 and from A/2 to A respectively. Then

$T_1/T_2$ will be equal to

0%
1
0%
1/2
0%
1/4
0%
2

For a wave propagating in a medium, identify the property that is independent of the others

0%
Velocity
0%
Wavelength
0%
Frequency
0%
All these depend on each other

A wave of frequency

$500Hz$ travels between X and Y, a distance of

$600 m$ in 2 sec. How many wavelength are there distance XY:-

0%
1000
0%
300
0%
180
0%
2000

Explanation

$\text{Step1:Finding Velocity of the wave }$

We have to calculate the velocity(V) of the wave by using the simple formula of velocity which is basically rate of change of the position respect to the time.

$\text{V=}{\dfrac{Displacement}{Time}}$

$={\dfrac{600m}{2sec}}$

$=300 \dfrac{m}{sec}$

[Given, the wave is travelling from point 'X' to point 'Y' having a distance of 600 m. So the displacement is 600 m. and the time taken by the wave is 2 sec.]

$\text{Step2:Finding the wavelength}$

The wavelength(

$\lambda$ ) is equals to the speed of the wave(V) divided by its frequency(f)

$\lambda=\dfrac{V}{f}=\dfrac{300m/sec}{500Hz}=\dfrac{3}{5}m$

$\text{Step3:Calculating number of waves}$

Number of waves can be calculated by the simple formula

$\text{Number of waves in XY displacement}=$

$\dfrac{Displacement}{\lambda}$

$=600m/(3/5)m=1000$

Hence, there are 1000 wavelength in distance XY.

A uniform rope of length

$L$ and mass

${m_1}$ hangs vertically from a rigid support . A block of mass

${m_2}$ is attached to the free end of the rope. A transverse pulse of wavelength

${\lambda _1}$ is produced at the lower end of the rope . the wavelength of the pulse when it reaches the top of the rope is

${\lambda _2}$ . The ratio

$\frac{{{\lambda _1}}}{{{\lambda _2}}}$ is:

0%
$\sqrt {\dfrac{{{m_1}}}{{{m_2}}}}$
0%
$\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_2}}}}$
0%
$\sqrt {\dfrac{{{m_2}}}{{{m_1}}}}$
0%
$\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_1}}}}$

A particle of mass

$m$ oscillates along

$x-$ axis according to equations

$x=a\sin \omega t$ . The nature of the graph between momentum and displacement of the particle is

0%
Straight line passing through origin
0%
Circle
0%
Hyperbola
0%
Ellipse

Two Waves of amplitudes

${ A }_{ 0 }$ and

$x{ A }_{ 0 }$ pass through a region. If x >1, the difference in the maximum and minimum resultant amplitude possible is

0%
$(x+1){ A }_{ 0 }$
0%
$(x-1){ A }_{ 0 }$
0%
$2x{ A }_{ 0 }$
0%
$2{ A }_{ 0 }$

Two particles execute

$SHM$ of same amplitude of

$20cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is

$20cm$ . Their phase difference in radians is :-

0%
$\cfrac{2\pi}{3}$
0%
$\cfrac{\pi}{2}$
0%
$\cfrac{\pi}{3}$
0%
$\cfrac{\pi}{4}$

Explanation

we have,

$\frac{{d\left| {{{\overline r }_{12}}} \right|}}{{dt}} = 0$

Rel. velocity=0

vel. same in 2nd and 3rd quad.

$\begin{array}{l} A\sin \theta =10 \\ \theta =\pi /6 \\ 2\theta =\pi /3 \end{array}$

Phase diff b/w particle.

So,

Option

$C$ is correct answer.

1215000_1248359_ans_bf2dc8a9278246189481b2e161a70093.JPG

A metal rod 40 cm long is dropped onto a wooden floor and rebounds into air. Compressional waves of many frequencies are thereby set up in the rod. If the speed of compressional waves in the rod is 5500

$m s ^ { - 1 }$ what is the - lowest frequency of compressional waves to which the rod resonates as it rebounds?

0%
6875 Hz
0%
16875 Hz
0%
675 Hz
0%
0.11 Hz

Explanation

$L = 2\left( {\frac{\lambda }{4}} \right)$

$,$

$\lambda = 2L = 2\left( {0.40m} \right) = 0.80m$

then

$,$ from

$\lambda = VT = \frac{V}{F}$

$F = \dfrac{V}{\lambda } = \dfrac{{5500}}{{0.80}} = 6875\,Hz$

Hence,

option

$(A)$ is correct answer.

1225808_1266512_ans_8b5f1c53f79b4869b0666d1d8dea40fc.png

Consider the wave represented by

$y=\cos(500t-70x)$ where

$x$ is in metres and

$t$ in seconds. the two nearest points in the same phase have a separation of

0%
$2\pi/7\ m$
0%
$2\pi/7\ cm$
0%
$20\pi/7\ m$
0%
$20\pi/7\ cm$

A particle is executing simple harmonic motion between extreme positions given by (-1, -2, -3)cm and (1, 2, 1)cm. Its amplitude of oscillation is

0%
6 cm
0%
4 cm
0%
2 cm
0%
3 cm

Two simple harmonic motions are represented by equations

$y _ { 1 } = 10 \sin ( 3 \pi t + \pi / 4 )$ and

$y _ { 2 } = 5 ( \sin 3 \pi t + \sqrt { 3 } \cos 3 \pi t ) \times 2$
Their amplitudes are in the ratio of

0%
2
0%
1
0%
3
0%
4

The equation of a stationary wave is

$y=0.8cos\left( \dfrac { \pi x }{ 20 } \right) \sin 200\pi t$ where x is in cm and t is in seconds. The separation between consecutive nodes is

0%
10 cm
0%
20 cm
0%
30 cm
0%
40 cm

The displacement of a particle is given by x = 3 sin

$\left( 5\pi t \right)$ + 4 cos

$\left( 5\pi t \right)$ The amplitude of particle is

0%
3
0%
4
0%
5
0%
7

CBSE Questions for Class 11 Engineering Physics Waves Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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