MCQ Questions for Class 11 Engineering Physics Waves Quiz 11

The displacement of an oscillating particle varies with time (in seconds) according to the equations

$y(cm)=sin\dfrac { \pi }{ 2} (\dfrac { t }{ 2}+\dfrac { 1 }{ 3} )$ . The maximum acceleration of the particle is approximately.

0%
$5.21cm/s^2$
0%
$3.62cm/s^2$
0%
$1.81cm/s^2$
0%
$0.62cm/s^2$

Explanation

From the given equation of the oscillation, we obtain,

The angular frequency of the oscillation

$\omega =\dfrac{\pi}{4}$

The amplitude of the particle is

$a=1$

So, the maximum acceleration of the particle under oscillation is given as:

$a_{max}=\omega^2 a =\left(\dfrac{\pi}{4}\right)^2 \times1$

$=0.62\ cm/sec^2$ [

$\because a=1$ ]

A tunnel is dug n the earth across one of its diameter. Two masses 'm' & '2m' are dropped from the ends of the tunnel. The masses collide and stick to each other and performs S.H.M. Then amplitude of S.H.M. will be : [R=radius of the earth]

0%
R
0%
R/2
0%
$\dfrac{{\sqrt {2R} }}{3}$
0%
2R/3

Explanation

For mass

$'m'$

$\begin{array}{l} ui+ki=uf+kf \\ \Rightarrow \dfrac { { -GMm } }{ R } +0=\dfrac { { -GM } }{ { 2{ R^{ 3 } } } } \left( { 3{ R^{ 2 } }-0 } \right) m+\dfrac { 1 }{ 2 } m{ v^{ 2 } } \\ \Rightarrow \dfrac { { -GMm } }{ R } =\dfrac { { -3GM } }{ { 2R } } +\dfrac { { { V^{ 2 } } } }{ 2 } \\ \Rightarrow \dfrac { { -GM } }{ R } =\dfrac { { { V^{ 2 } } } }{ 2 } \\ \Rightarrow { V^{ 2 } }=\dfrac { { 2GM } }{ R } \\ \Rightarrow V=\sqrt { \dfrac { { 2GM } }{ R } } \end{array}$

Conserving momentum before and after collinear.

Let

${V_{cm}}$ is the velocity of combines mass.

${V_{cm}} = \dfrac{{2mv - mv}}{{3m}} = \dfrac{v}{3}$

Let the amplitude is

$x$ .

$\begin{array}{l} ui+ki=uf+kf \\ \Rightarrow \dfrac { { -3GM\left( { 3m } \right) } }{ { 2R } } +\dfrac { 1 }{ 2 } \times 3m\times \dfrac { { { v^{ 2 } } } }{ 9 } =\dfrac { { -GM } }{ { 2{ R^{ 3 } } } } \left( { 3{ R^{ 2 } }-{ x^{ 2 } } } \right) \left( { 3m } \right) +0 \\ \Rightarrow x=\dfrac { { \sqrt { 2R } } }{ 3 } \end{array}$

$\therefore$ AMplitude is

$\dfrac{{\sqrt {2R} }}{3}$

$\therefore$ Option

$C$ is correct answer.

1207894_1281132_ans_3d4ed2ce9df942cb8410e4c954088ec9.JPG

Equation of motion in the same direction is given by

$y_1 = A \ sin(\omega t - kx), y_2 = A \ sin(\omega t - kx - \theta)$ . The amplitude of the medium particle will be

0%
2 A $cos\frac{\theta}{2}$
0%
2 A $cos \theta$
0%
A $cos \theta$
0%
A $cos\frac{\theta}{2}$

A body executing simple harmonic motion has a maximum acceleration equal to

$24metre/sec^2$ and maximum velocity equal to

$16meter/sec$ . The amplitude of simple harmonic motion is:

0%
$\dfrac { 2 }{ 32} m$
0%
$\dfrac { 32}{ 3}m$
0%
$(\dfrac { 64 }{ 9 } )m$
0%
$(\dfrac { 1024 }{ 9 } )m$

The graph shown the variation of displacement of a particle executing SHM with time. We inference from this graph that-

0%
the force is zero at time $\dfrac{3T}{4}$
0%
the velocity is maximum at time $\dfrac{T}{2}$
0%
the acceleration is maximum at time $T$
0%
the P.E. is equal to half of total energy at time $\dfrac{T}{2}$

Explanation

As the graph shows the dissplacement vs time for a particle performing SHM,

At time

$\dfrac T2$ , the displacem,ent of particle is zero. So, it will be at the mean position of its SHM.

At time

$\dfrac{T}{2};v=max$

$\therefore$ Total energy will be stored in form of the kinetic energy.

So, at

$\dfrac T2$ , the velocity will be maximum.

A travelling wave

$Y = A$ sin

$( k x - \omega t + \theta )$ passes from a heavier string to lighter string. The reflected has amplitude

$0.5$

$A$ . The junction of the string is

$x = 0$ . The equation of the reflected wave is:

0%
$y ^ { \prime } = 0.5 A \sin ( k x + \omega t + \theta )$
0%
$y ^ { \prime } = - 0.5 A \sin ( k x + \omega t + \theta )$
0%
$y ^ { \prime } = - 0.5 A \sin ( \omega t - k x - \theta )$
0%
$y ^ { \prime } = - 0.5 A \sin ( k x + \omega t - \theta )$

Explanation

$\begin{array}{l} Y=A\sin \left( { Kx-\omega t+\theta } \right) \\ y'=0.5A\sin \left( { Kx+\omega t+\theta +\pi } \right) \\ =-0.5A\sin \left( { Kx+\omega t+\theta } \right) \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$

The amplitude of simple harmonic motion represented by the displacement equation y(cm) = 4(sin 5

$\pi$ t +

$\sqrt2$ cos 5

$\pi$ t) is :

0%
4 cm
0%
4 $\sqrt2$ cm
0%
4 $\sqrt3$ cm
0%
4( $\sqrt2 + 1) cm$

Explanation

$y = u (sin5\pi t+\sqrt{2}cos5\pi t)$

$=u sin 5 \pi zt+4\sqrt{2}cos5\pi t$

$y = A sinwt + B coswt$

amplitude

$= \sqrt{A^{2}+B^{2}}$

$= \sqrt{(4)^{2}+(4\sqrt{2})^{2}}$

$= 4\sqrt{3}$

Option (c)

1166255_1324285_ans_a8e7a7cb585c4281ac914aac002331f1.jpg

$4kg$ particle is moving along the x-axis under the action of the force which varies with distance

$x$ given by

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

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

$t=10sec$ , its speed is

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

0%
$\dfrac { 32\sqrt { 2 } }{ \pi } m$
0%
$\dfrac { 16 }{ \pi } m$
0%
$\dfrac { 4 }{ \pi } m$
0%
$\dfrac { 64\sqrt { 2 } }{ \pi } m$

The equation of wave is given by $y = 10 \sin(2\pi t/45 + \alpha)$ . If the displacement is $5 cm$ at $=0$ , then the total phase at $t = 7.5 s$ will be:

0%
$\pi/3$ rad
0%
$\pi/2$ rad
0%
$2\pi/5$ rad
0%
$2\pi/3$ rad

What is the period of a vibrating particle executing S.H.M.. Which has acceleration

$0.12 m/s^2$ when its displacement is

$3 \times 10^{-2} m?$

0%
2 s
0%
1 s
0%
$2 \pi s$
0%
$\pi s$

Two waves represented as

$y_1 = a \sin (\omega t + {\pi}/6)$ ,

$y_2 = a \cos \omega t$
The resultant amplitude is-

0%
$a$
0%
$a\sqrt { 2 }$
0%
$a\sqrt {3}$
0%
$2a$

The amplitude of a particle executing

$SHM$ is

$4cm$ . At the mean position the speed of the particle is

$16cm/sec$ . The distance of the particle from the mean position at which the speed of the particle becomes

$8\sqrt { 3} cm/s$ , will be:

0%
$2\sqrt { 3} cm/s$
0%
$\sqrt { 3} cm/s$
0%
$1cm$
0%
$2cm$

Explanation

At mean position velocity is maximum

i.e.,

$v_{max}=\omega a \Rightarrow \omega =\dfrac{v_{max}}{a}=\dfrac{16}{4}=4$

The velocity can also be represented as:

$\therefore v=\omega \sqrt{a^2-y^2}$

$\Rightarrow 8\sqrt 3=4\sqrt{4^2-y^2}$

$\Rightarrow 192=16(16-y^2)\Rightarrow 12=16-y^2$

$\Rightarrow y=2\ cm$ .

A particle executes simple harmonic motion and is located x = a, b and c at time

$\ t_0 \ 2t_0 \ and \ 3t_0$ respectively.The frequency of the oscillation is:-

0%
$\cfrac { 1 }{ 2\pi t_ 0 } cos^ {-1} \left( \frac { a+b }{ 2c } \right)$
0%
$\dfrac { 1 }{ 2\pi t_ 0 } cos^ {-1} \left( \frac { a+b }{ c } \right)$
0%
$\dfrac { 1 }{ 2\pi t_ 0 } cos^ {-1}\left( \frac { a+2b }{ 3c } \right)$
0%
$\dfrac { 1 }{ 2\pi t_ 0 } cos^{- 1}\left( \frac { a+c }{ 2b } \right)$

On the superposition of the two waves given as

${ y }_{ 1 }={ A }_{ 0 }sin\left( \omega t-kx \right)$ and

${ y }_{ 2 }={ A }_{ 0 }cos\left( \omega t-kx+\dfrac { \pi }{ 6 } \right)$ . the resultant amplitude of oscillations will be

0%
$\sqrt { 3 } { A }_{ 0 }$
0%
$\dfrac { { A }_{ 0 } }{ 2 }$
0%
${ A }_{ 0 }$
0%
$\dfrac { 3 }{ 2 } { A }_{ 0 }$

$If x=a\sin { (\omega +\frac { \pi }{ 6} })$ and

$x^2=a\cos \omega { t }$ ,then what is the phase difference between the two waves:

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{2}$
0%
$\pi$

Two waves are represented by the equations

$y_1 = a sin (\omega + kx + 0.57 )m$ and

$y_2 = a cos (\omega t + kx)m$
where x is in meter and t in sec. The phase difference between them is

0%
0.57 radian
0%
1.0 radian
0%
1,25 radian
0%
1.57 radian

Explanation

$y_{1}=A sin (\omega t+kx+0.57)$

$y_{2}=A cos (\omega t+kx)$

$= A sin (\omega t+kx+\frac{\pi }{2})$

phase difference =

$\frac{\pi }{2}-0.57$

$=1.57-0.57$

= 1 radian

Option (B)

1166849_1338730_ans_84ece092cd894ab7a4c203e845eaa788.jpg

Two particles undergo SHM along the same line with the same time period (

TT They will cross each other after a further time

0%
$t = \dfrac{{T}}{2}$
0%
$t = \dfrac{T}{\sqrt {2}}$
0%
$t= \dfrac{2T}{4}$
0%
$t = \dfrac{3T}{8}$

Explanation

Hence, option

$(D)$ is correct answer.
1380706_1345664_ans_07a758ce538047e99a8b57c2e8eb9578.jpg

Two waves whose intensity are same

$I$ move towards a point

$P$ in same phase, then the resultant intensity at point

$P$ will be :

0%
$4\, I$
0%
$2\, I$
0%
$\sqrt{2}\, I$
0%
$\sqrt{6}\, I$

A particle performs SHM with amp

$A$ . Its speed is hetled when at

$\dfrac {2A}{3}$ from equation. Two new amp is

0%
$3A$
0%
$A\sqrt {3}$
0%
$\dfrac {7A}{3}$
0%
$\dfrac {A}{3}\sqrt {41}$

Equation of progressive wave is given by

$y=a\sin \pi { (\frac {t }{ 2}- \frac { x }{4 } ) }$ ,where t is in seconds and x is in meter. Then the distance through which the wave moves in 4 seconds is (in meter)-

0%
$4m$
0%
$2m$
0%
$16m$
0%
$8m$

Explanation

y=a sin

$\pi (\frac{t}{2}-\frac{x}{4})$

Compare this with standard equation, y=A sin 2

$\pi (\frac{t}{T}-\frac{x}{\lambda })$

where, A= amplitude

$\lambda$ = wavelength = 4

T= time period = 2

$\frac{1}{T}=\frac{1}{2}= 0.5$

f= frequency

$f=\frac{1}{T}= 0.5$

f= 0.5 Hz

Wave velocity,

$v=f\times \lambda$

$v=0.5\times 4=2 cm/s$

1172010_1334397_ans_ac8bde745ef3409d9052cb42ceec005f.JPG

Two wave are represnted by the equations

$y_1= a sin \omega t \ and \ y_2 = a cos \omega t$ . the first wave :

0%
leads the second by $\pi$
0%
lags the second by $\pi$
0%
leads the second by $\frac { \pi}{2}$
0%
lags the second by $\frac { \pi}{2}$

Explanation

Given,

$y_1 = a\sin \omega t$

$y_2 = a\cos\omega t$

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

So,

$y_1$ lags

$y_2$ by

$\pi /2$ .

So, the correct option will be

$(D)$

Three coherent waves having amplitudes 12 mm, 6 mm and 4 mm arrive at a given point with successive phase difference of

$\frac{\pi}{2}$ . Then the amplitude of the resultant wave is:

0%
7 mm
0%
10 mm
0%
5 mm
0%
1.8 mm

Explanation

From the vector diagram,

$(A_{net})_{13}=\sqrt{A_1^1+A_3^2+2A_1A_3cos\delta}$

$(A_{net})_{13}=\sqrt{(12)^2+(4)^2+2\times 12\times 4\times cos180^0}=8mm$

Now the net amplitude,

$A_n=\sqrt{A_{13}^2+A_2^2+2A_2A_{13}cos90^0}=\sqrt{A_{13}^2+A_2^2}$

$A_n=\sqrt{(6)^2+(8)^2}=10mm$

The correct option is B.

1480051_1384802_ans_50125672ee0a46a3a35244ab3677dfbe.png

The phase difference between two points separated 0.8 m in a wave of frequency 120 HZ is 0.5

$\pi$ the value velocity is

0%
144 ${ ms }^{ -1 }$
0%
384 ${ ms }^{ -1 }$
0%
256 ${ ms }^{ -1 }$
0%
720 ${ ms }^{ -1 }$

A particle is vibrating in simple harmonic motion with amplitude of

$4\ cm$ . At what displacement from the equilibrium position is its energy half potential and half kinetic?

0%
$1\ cm$
0%
$\sqrt {2}\ cm$
0%
$2\ cm$
0%
$2\sqrt {2}\ cm$

Explanation

$Kinetic\,\,energy = \frac{1}{2}m{w^2}\left( {{A^2} - {x^2}} \right)$

Where x is the displacement from mean position

$\begin{array}{l} Potential\, \, energy=\frac { 1 }{ 2 } m{ w^{ 2 } }{ x^{ 2 } } \\ \frac { 1 }{ 2 } m{ w^{ 2 } }{ x^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }\left( { { A^{ 2 } }-{ x^{ 2 } } } \right) \\ { x^{ 2 } }={ A^{ 2 } }-{ x^{ 2 } } \\ { x^{ 2 } }-{ x^{ 2 } }.{ A^{ 2 } } \\ 2{ x^{ 2 } }={ 4^{ 2 } } \\ 2{ x^{ 2 } }=16 \\ { x^{ 2 } }=8 \\ x=2\sqrt { 2 } \, \, cm \end{array}$

Option D.

Two particles in the path of a wave, of velocity

$360\ m/s$ and frequency

$2000\ Hz$ , differ in phase by

$120^{o}$ . The distance between the particles is

0%
$6\ cm$
0%
$12\ cm$
0%
$18\ cm$
0%
$100\ cm$

Displacement of particle along x-axis by

$x=a\sin^{2}{\omega t}$ . Its motion is SHM of frequency :

0%
$\omega /\pi$
0%
$3\omega /2\pi$
0%
$\omega /2\pi$
0%
$not\ SHM$

Explanation

We have,

$x= a\sin^2 \omega t$

$\Rightarrow \dfrac{dx}{dt} = 2 \omega (\sin \omega t) ( \cos \omega t)$

$\Rightarrow \dfrac{d^2x}{dt^2} = 2 \omega (\cos ^2 \omega t - \sin^2 \omega t)$

$\Rightarrow a = 2a \omega^2\cos 2 \omega t$

But for SHM,

$a \propto - \omega^2$

So, This is not

$SHM$ .

So, the correct option will be

$(D)$

A particles starts its

$SHM$ from mean position at

$t = 0$ . If its time period is

$T$ and amplitude

$A$ . The distance travelled by the particle in the time from

$t = 0$ to

$t = \dfrac{5T}{4}$ is

0%
$A$
0%
$2A$
0%
$4A$
0%
$5A$

Two waves of equal amplitude when superposed , give a resultant wave having an amplitude equal to that of either wave. The phase difference between the two waves is

0%
$\dfrac { \pi }{ 3 } \quad raidan$
0%
zero
0%
$\dfrac { \pi }{ 2 } \quad raidan$
0%
$\dfrac { 2\pi }{ 3 } \quad raidan$

The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 0.5

$\pi$ the wave velocity is

0%
144 ${ ms }^{ -1 }$
0%
384 ${ ms }^{ -1 }$
0%
256 ${ ms }^{ -1 }$
0%
720 ${ ms }^{ -1 }$

Two SHM are represented by equations

${ y }_{ 1 }=6cos\left( 6\pi t+\dfrac { \pi }{ 6 } \right) ,{ y }_{ 2 }=3\left( \sqrt { 3 } sin3\pi t+cos3\pi t \right)$

0%
Ratio of their amplitudes is 1
0%
ratio of their time periods is 1
0%
ratio of their maximum velocities is 1
0%
ratio of their maximum acceleration is 1

Explanation

$\begin{array}{l} { y_{ 1 } }=6\cos \left( { 6\pi t+\frac { \pi }{ 6 } } \right) \\ { y_{ 2 } }=3\left( { \sqrt { 2 } \sin 3\pi t+\cos 3\pi t } \right) \\ =6\left[ { \frac { { \sqrt { 3 } } }{ 2 } \sin 3\pi t+\frac { 1 }{ 2 } \cos 3\pi t } \right] \\ 6\left[ { \sin \left( { 3\pi t+\frac { \pi }{ 3 } } \right) } \right] \\ =6\sin \left( { 3\pi t+\frac { \pi }{ 3 } } \right) \\ ratio\, \, of\, their\, amplitude\, is\, 1. \end{array}$

Hence,

Option

$A$ is correct answer.

Two particles undergo SHM along same line with same time period (T) and same amplitude (A) At particular instant one particle is x = - A and the other is at x = 0 If they move in same direction then they will meet each other at

0%
$x = \dfrac { A } { 2 }$
0%
$x = \dfrac { A } { \sqrt { 2 } }$
0%
$x = \dfrac { A } { 4 }$
0%
$x = \dfrac { 4 } { 8 }$

Explanation

Representing the SHM in ordinance motion.

As we see the projection of the balls meet at distance

$A\sin 45^o=\dfrac{A}{\sqrt{2}}$ .

1205984_1400288_ans_5fe819ba07754b38ba4be9f1cab92175.jpg

The speed of sound in a certain medium is

$960 m/s$ . If

$3600$ waves pass over a certain point in 1 minute, the wavelength is,

0%
$2 m$
0%
$8 m$
0%
$4 m$
0%
$16 m$

Explanation

$v=960m/s, \\ n=\dfrac{3600}{60}Hz = 60 Hz$

$λ = \dfrac{v}{n}=\dfrac{960}{60}=16m$

Hence option 'D' is correct.

Two particles execute S.H.M. of same amplitude and frequency along the same straight line. They pass one another, when going in opposite directions, each time their displacement is half of their amplitude. The phase-difference between them is

0%
$90 ^ { \circ }$
0%
$120 ^ { \circ }$
0%
$180 ^ { \circ }$
0%
$135 ^ { \circ }$

Explanation

$\begin{array}{l} For\, \, particle\, in\, S.H.M \\ y=A\, \sin wt\, \, phase=\phi =\omega t \\ y=\dfrac { A }{ { \sqrt { 2 } \, } } \, \, for\, { P_{ 1 } } \\ \dfrac { A }{ { \sqrt { 2 } } } =A\, \sin \omega t \\ \left( { \omega ,\, t } \right) ={ \phi _{ 1 } }=\dfrac { \pi }{ 4 } \, \, \, \, \, \, \, \, \, ----\left( 1 \right) \\ For\, \, { p_{ 2 } }-\, in\, opposite\, direction \\ phase\, { \phi _{ 2 } }=\left( { \pi -\dfrac { \pi }{ 4 } } \right) =\left( { \dfrac { { 3\pi } }{ 4 } } \right) \, \, \, \, \, \, ---\left( 2 \right) \\ \Delta \phi =\left( { { \phi _{ 2 } }-{ \phi _{ 1 } } } \right) =\dfrac { { 3\pi } }{ 4 } -\dfrac { \pi }{ 4 } =\dfrac { \pi }{ 2 } ={ 90^{ 0 } } \end{array}$

Hence, the option

$A$ the correct answer.

1234608_1406272_ans_f28ad47951184e68b9797518fcfaccb5.PNG

The displacement of a particle along the x axis is given by

$x=a{ sin }^{ 2 }\omega t.$ The motion of the particle corresponds to

0%
simple harmonic motion of frequency $\omega /\pi$
0%
simple harmonic motion of frequency $3\omega /2\pi$
0%
non simple harmonic motion
0%
simple harmonic motion of frequency $\omega /2\pi$

When a wave travels in a medium, the particle displacement is given by y(xt)=0.03 sin

$\pi$ (2t-0.01 x) where y and x are meters and t in seconds. The phase difference, at a given instant of time between two particle 25 m. apart in the medium, is

0%
$\frac{\pi }{8}$
0%
$\frac{\pi }{4}$
0%
$\frac{\pi }{2}$
0%
$\pi$

the amplitude and frequency of a wave represented by the equation

$Y= 2 A cos^2 (kx-wt)$ are :

0%
$2 A, \dfrac {\omega}{2 \pi}$
0%
$A, \dfrac {\omega}{ \pi}$
0%
$\sqrt A, \dfrac {\omega}{ \pi}$
0%
$\sqrt A, \dfrac {\omega}{2 \pi}$

The equation of

$SHM$ with amplitude

$4m$ and time period

$2sec$ with initial phase

$\dfrac{\pi}{3}$ is

$y =$

0%
$4sin(2\pi t+\dfrac{\pi}{3})m$
0%
$4sin(\pi t+\dfrac{\pi}{3})m$
0%
$4sin(4\pi t+\dfrac{\pi}{3})m$
0%
$4sin(\pi t+\dfrac{\pi}{6})m$

The figure shown an instantaneous profile of a rope carrying a progressive mave moving from left to right, then :-
(A) The phase at A is greater then the phase at B
(B) The phase at B is greater then the phase at A
(C) A is moving upwards
(D) B is moving upwards

0%
(A) and (C)
0%
(A) and (D)
0%
(B) and (C)
0%
(B) and (D)

A particle performs simple harmonic motionwith amplitude A. Its speed is trebled at the instant when it is at a distance

$\dfrac { 2 \mathrm { A } } { 3 }$ fromequilibrium position. Find the new amplitudeof the motion.

0%
$\dfrac { 7 \mathrm { A } } { 3 }$
0%
$\dfrac { \mathrm { A } } { 3 } \sqrt { 41 }$
0%
3 $\mathrm { A }$
0%
$\mathrm { A } \sqrt { 3 }$

A sound wave travels at a speed of

$330m/s.$ If the wavelength is

$1.56cm$ what is the frequency of the wave? will it be audible for human

0%
$50Hz$
0%
$30Hz$
0%
$20Hz$
0%
$45Hz$

Explanation

$s=330$

$h=1.5cm=1.5/100=0.015$

$s×h$

$330×0.015$

$=4.95 Hz$

it Will not be audible as it less than $20Hz$

Hence,

option $(C)$ is correct answer.

A particle in S.H.M. is described by the displacement function

$x(t) = a cos(ax + \theta)$ . If the initial (t = 0) position of the particle is 1 cm and its initial velocity is

$\pi cm/s$ . The angular frequency of the particle is

$\pi rad /s$ ,then it's amplitude is

0%
1 cm
0%
$\sqrt{2}cm$
0%
2 cm
0%
2.5 cm

Explanation

$\begin{array}{l} Given, \\ Diplacement\, \, function\, \, of\, \, the\, \, particle\, \, is, \\ x\left( t \right) =A\cos \left( { \omega t+\phi } \right) \\ x\left( 0 \right) =1\, \, cm\, \, and\, \, \, v\left( 0 \right) \, \, =\pi \, \, \dfrac { { cm } }{ { \sec } } \\ differentiating\, \, w.r.\, \, to\, \, t, \\ v\left( t \right) =-A\left( \omega \right) \sin \left( { \omega t+\varphi } \right) ..........\left( 1 \right) \\ At\, \, t=0, \\ x\left( 0 \right) =A\omega \sin \left( \phi \right) ..........\left( 2 \right) \\ v\left( 0 \right) =-A\omega \sin \left( \phi \right) ............\left( 3 \right) \\ Dividing\, \, \left( 3 \right) \, \, by\, \, \left( 2 \right) , \\ \dfrac { { v\left( 0 \right) } }{ { x\left( 0 \right) } } =-\omega \tan \phi \\ \Rightarrow \pi =-\omega \tan \phi \\ \Rightarrow \tan \phi =-1 \\ i.e\, \, \, \phi =\dfrac { { 3\pi } }{ 4 } \, \, or\, \, \dfrac { { 7\pi } }{ 4 } \\ \sin ce\, \, ,at\, \, \, t=0 \\ x\left( 0 \right) =A\cos \left( \phi \right) =1 \\ \Rightarrow \cos \phi \, \, \, is\, \, \, positive. \\ Therefore\, \, ,\phi =\dfrac { { 7\pi } }{ 4 } \\ \therefore A\cos \left( { \dfrac { { 7\pi } }{ 4 } } \right) =1 \\ \Rightarrow A=\sqrt { 2 } \, \, cm \end{array}$

The phase difference between two waves represented by

${ y }_{ 1 }={ 10 }^{ -6 }sin\left[ 100t+\left( \times /50 \right) \right] m$

${ y }_{ 2 }={ 10 }^{ -6 }cos\left[ 100t+\left( \times /50 \right) \right] m$
where x is expressed in meters and t is expressed in seconds, is approximately

0%
1.07 rad
0%
2.07 rad
0%
0.0 rad
0%
1.5 rad

Consider the following two equations

$L=I\omega$ and

$\dfrac { dL }{ dt } =\Gamma$ . In noninertial frames :

0%
both A and B are true
0%
A is true but B is false
0%
B is true but A is false
0%
both A and B are false.

Explanation

Angular momentum

$L = Iw$

A rigid object's angular momentum is stated as follows :

The cross product of the moment of inertia and angular velocity.
It is considered as analogous to corresponding linear momentum.
If there are no other external torque acts on the object, it is subject to the basic limitations of the angular momentum theory.

$\dfrac{dL}{dt}$ = Total external force torque not equal to any torque.

So,

$\dfrac{dL}{dt}$ = Γ is not correct.

Hence, option (b) A is true but B is false.

If x=

$\theta sin(\alpha + \dfrac{\pi}{6})$ and

$x^1 = {\theta}cos\alpha$ ,then what is the phase difference between the two waves.

0%
$\pi/3$
0%
$\pi/6$
0%
$\pi/2$
0%
$\pi$

The displacement of an object attached to a spring and executing simple harmonic motion is given by

$x=2\times { 10 }^{ -2 }\cos { \pi t }$ meters. The time at which the maximum speed first occurs is :

0%
$0.5 s$
0%
$0.75 s$
0%
$0.125 s$
0%
$0.25 s$

A boat has green light of wavelength

$\lambda= 500nm$ on the MAST. What wavelength would be measured and what colour would be observed for this light as seen by a diver submerged in water by the side of the boat?

Given

$\eta_{w} = 4/3$

0%
Green of wavelength $500 nm$
0%
Blue of wavelength $376nm$
0%
Green of wavelength $375 nm$
0%
Red of wavelength $665 nm$

Speed of wave 'v' is given by

0%
Wavelength of the wave / frequency of the wave
0%
Wavelength of the wave $\times$ frequency of the wave
0%
frequency of the wave/Wavelength of the wave
0%
None of these

For a wave displacement amplitude is

$10^-4 m$ ,density of air 1.3 kg

$m^-4$ ,velocity in air 340

$ms^-4$ are frequency is 2000 Hz.The intensity of wave is

0%
$5.3\times 10^-4 Wm^-4$
0%
$5.3\times 10^4 Wm^-4$
0%
$3.5\times 10^-4 Wm^4$
0%
none of these

Explanation

$\begin{array}{l} From\, the\, question \\ A={ 10^{ -8 } }m \\ \rho =1.3\, kg/{ m^{ 3 } } \\ v=340\, m/s \\ v=2000\, Hz \\ The\, \, { { int } }ensity\, of\, wave\, \, is\, given\, by\, the\, following\, relation \\ I=\frac { 1 }{ 2 } \rho { A^{ 2 } }{ \omega ^{ 2 } }c\, \, \, \, \, \, \, \, \, \, ---\left( 1 \right) \\ \omega \, is\, angular\, frequency\, ,\, \omega =2\pi v \\ c\, is\, the\, speed\, of\, light \\ So,\, equation\, \left( 1 \right) \, becomes \\ I=\frac { 1 }{ 2 } \times 1.3\, \, kg/{ m^{ 3 } }\times { \left( { { { 10 }^{ -8 } }m } \right) ^{ 2 } }\times { \left( { 2\pi \times 2000\, Hz } \right) ^{ 2 } }\times 3\times { 10^{ 8 } }m/s \\ I=3.0761\, W/{ m^{ 2 } } \end{array}$

Hence, the option

$D$ is the correct answer.

The S.H.M. of a particle is given by the equation

$y = 3 \sin \omega t + 4 \cos \omega t .$ The amplitude is

0%
7
0%
1
0%
5
0%
12

Four waves of the equation given by

$y_1= 5A \ sin \ (wt - Kx + \pi/2)$

$y_2= 2A \ sin \ (wt - Kx + 3 \pi/2)$

$y_3= 6A \ sin \ (wt - Kx)$
and

$y_4= 2A \ sin \ (wt - Kx + \pi)$

0%
the resultant wave will have amplitude 5A.
0%
It will make a phase difference of $37^o$ with $\gamma_1$ .
0%
Its equation will be $y=5A \ sin \ (wt + 37^o).$
0%
Its equation will be $y=5A \ sin \ (wt -Kx + 37^o).$

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.