MCQ Questions for Class 11 Engineering Physics Waves Quiz 8

Two particles move parallel to

$x$ -axis about the origin with same amplitude and frequency. At a certain instant they are found at a distance

$\dfrac{a}{3}$ from the origin on opposite sides but their velocities are found to be in same direction.Then the phase difference between two particles will be

0%
$cos^{-1}( \dfrac{7}{9})$
0%
$180^{0}$
0%
$45^{0}$
0%
None of the above

Explanation

Let phase difference between them is

$\phi$ .
So,

$x_1=a\,sin \,\omega t$ and

$x_2=9\,sin (\omega t+\phi)$
or

$\dfrac{a}{3}=a\,sin\,\omega t$

$\Rightarrow$

$sin \,\omega t =\dfrac{1}{3}$ ...............(i)
Similarly ,

$-\dfrac{a}{3}= a\,sin\,(\omega t+\phi)$

$sin\,(\omega t +\phi)=-\dfrac{1}{3}$ ...................................(ii)
Eliminating

$t,$ we get

$cos \,\phi=-1\,,\dfrac{7}{9}$

$\Rightarrow$

$\phi=180^{0}$
or

$\phi=cos^{-1}\,(\dfrac{7}{9})$

The ratio of amplitudes of two simple harmonic motions represented by the equations

$y_1=5 \sin \begin{pmatrix}2\pi t + \dfrac{\pi}{4}\end{pmatrix}$ and

$y_2=2\sqrt{2} (\sin 2\pi t + \cos 1\pi t)$ is

0%
1 : 1
0%
2 : 1
0%
5 : 2
0%
5 : 4
0%
2 : 5

Explanation

The amplitude for first SHM equation,

$A_1=\sqrt{5^2}=5$ and

the amplitude for second SHM equation,

$A_2=\sqrt{(2\sqrt 2)^2+(2\sqrt 2)^2}=\sqrt{8+8}=\sqrt {16}=4$

Thus,

$\dfrac{A_1}{A_2}=\dfrac{5}{4}$

With propagation of longitudinal waves through a medium, the quantity transmitted is :

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matter
0%
energy
0%
energy and matter
0%
energy, matter and momentum

Two simple harmonic motions are represented by

${ y }_{ 1 }=5\left( \sin { 2\pi t } +\sqrt { 3 } \cos { 2\pi t } \right)$

${ y }_{ 2 }=5\left( \sin { 2\pi t } +\cfrac { \pi }{ 4 } \right)$
The ratio of the amplitude of two S.H.M's is

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:\sqrt { 3 }$

If the equation

$y_{1} = A\sin \omega t$ and

$y_{2} = \dfrac {A}{2}\sin \omega t + \dfrac {A}{2}\cos \omega t$ represent SHM then the ratio of the amplitudes of the two motions is

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

Explanation

Given,

$y_{1} = A\sin \omega t$
and

$y_{2} = \dfrac {A}{2} \sin \omega t + \dfrac {A}{2} \cos \omega t$
or

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

$= \dfrac {A}{3} \sqrt {2} [\sin (\omega t + 45^{\circ}]$

$= \dfrac {A}{\sqrt {2}} \sin (\omega t + 45^{\circ})$

$\therefore$ The ration of amplitudes of two motions

$\dfrac {A_{1}}{A_{2}} = \dfrac {A}{A\sqrt {2}} = \sqrt {2}$ .

The equation of simple harmonic wave is given by

$y=6 \,sin \,2\pi ( 2 t-0.1 x),$ where

$x$ and

$y$ are in mm and

$t$ is in seconds . The phase difference between two particles

$2 mm$ apart at any instant is

0%
$18^{0}$
0%
$36^{0}$
0%
$54^{0}$
0%
$72^{0}$

Explanation

Given equation can be written as

$y = 6\sin(4\pi t- 0.2\pi x)$

$\therefore$ Phase

$\phi = -0.2\pi x$

$\Rightarrow\phi_1 = -0.2\pi x_1$ and

$\phi_2 = -0.2\pi x_2$

$\therefore$ Phase difference

$\Delta\phi =|\phi_2 - \phi_1| =| -0.2\pi(x_2-x_1)|$

given

$(x_2 - x_1) = 2mm$ substituting in above equation we get

$\Delta\phi = 0.2\times{180^0}\times{2} = 72^0$

In a ripple tank when one pulse is sent every tenth of a second , the distance between consecutive pulses is

$30 mm$ . In the same depth of water pulses are produced at half second intervals. What is the new distance between consecutive pulses ?

0%
$0.67 mm$
0%
$6.0 mm$
0%
$150 mm$
0%
$600 mm$

Explanation

At the same depth under water velocity of sound will be same

$\therefore$

$v_1=v_2$

$\Rightarrow$

$\dfrac{\lambda_1}{T_1}=\dfrac{\lambda_2}{T_2}$

$\therefore$

$\lambda_2=\dfrac{T_2}{T_1}\lambda_1$

$=\dfrac{1/2}{1/10} \times 30 =150 mm$

Five waveforms moving with equal speeds on the x-axis

${ y }_{ 1 }=8\sin { \left( \omega t+kx \right) } ;{ y }_{ 2 }=6\sin { \left( \omega t+\cfrac { \pi }{ 2 } +kx \right) } ;{ y }_{ 3 }=4\sin { \left( \omega t+\pi +kx \right) } ;$

${ y }_{ 4 }=2\sin { \left( \omega t+\cfrac { 3\pi }{ 2 } +kx \right) } ;$

${ y }_{ 5 }=4\sqrt { 2 } \sin { \left( \omega t-kx+\cfrac { \pi }{ 4 } \right) }$ are superimposed on each other. The resulting wave is:

0%
$8\sqrt { 2 } \cos { kx } \sin { \left( \omega t+\cfrac { \pi }{ 4 } \right) }$
0%
$8\sqrt { 2 } \sin { \left( \omega t-kx+\cfrac { \pi }{ 4 } \right) \quad \quad }$
0%
$8\sqrt { 2 } \sin { kx } \cos { \left( \omega t+\cfrac { \pi }{ 4 } \right) }$
0%
$8\sin { \left( \omega t+kx \right) }$

Explanation

Lets first find resultant of four waves using phase diagram

first wave is representated along $+ve$ X-axis with vector of length equal to amplitude of the wave.

resultant of those four can be written as (by adjusting X and Y component)

$R=\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =4\sqrt { 2 }$

and $\sin { \theta =\cfrac { 4 }{ 4\sqrt { 2 } } } =\cfrac { 1 }{ \sqrt { 2 } } \\ \therefore \theta =\cfrac { \pi }{ 4 } \\ \therefore { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }=4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi }{ 4 } \right) }$

now add fifth wave

$\therefore y=\left( { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } \right) +{ y }_{ 5 }\\ =4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi }{ 4 } \right) } +4\sqrt { 2 } \sin { \left( wt-kx+\cfrac { \pi }{ 4 } \right) } \\ =4\sqrt { 2 } \left[ 2\times \sin { \left( wt+\cfrac { \pi }{ 4 } \right) cos\left( kx \right) } \right]$

By using $\sin { C } +\sin { D } =2\sin { \cfrac { C+D }{ 2 } \cos { \cfrac { C+D }{ 2 } } }$

$\therefore y=8\sqrt { 2 } \sin { \left( wt+\cfrac { \pi }{ 4 } \right) } \cos { \left( kx \right) }$

1021835_764591_ans_820edb6d13ee491684d687b5619ddffc.png

Equations of a stationary wave and a travelling wave are

$y_1=1\,sin(kx)\,cos (\omega t)$ and

$y_2=a\,sin\,(\omega t-kx)$ .The phase difference between two points

$x_1=\dfrac{\pi}{3k}$ and

$x_2=\dfrac{3 \pi}{2k}$ is

$\phi_1$ for the first wave and

$\phi_2$ for the second wave.The ratio

$\dfrac{\phi_1}{\phi_2}$ is

0%
1
0%
5/6
0%
3/4
0%
6/7

Explanation

Phase difference between two points in a standing wave =

$n\pi$

Where n is number of nodes between two points.

Given points are

$x_1 = \cfrac{\pi}{3k} = \cfrac{60}{k}$

$x_2 = \cfrac{3\pi}{2k} = \cfrac{210}{k}$

Equation of the standing wave

$y_1 = a \sin kx \cos \omega t$

At node points

$kx =n\pi$

$x = \cfrac{n\pi}{k} \quad (n=0,1,2,3...)$

So nodes are =

$\cfrac{\pi}{k} , \cfrac{2\pi}{k} ....$

$= \cfrac{180}{k} , \cfrac{360}{k} ....$

Since there is only one node between phase difference

$\phi_1 = \pi$

For travelling wave

$\phi _2 = \cfrac{2\pi}{\lambda} \triangle x$

From the equation

$y_2 = a \sin (\omega t - kx)$

$k = \cfrac{2\pi}{\lambda}$

$\therefore \phi_2 = k[x_2 - x_1] = k[\cfrac{3\pi}{2k} - \cfrac{\pi}{3k}] = \cfrac{7}{6}\pi$

$\therefore \cfrac{\phi_1}{\phi_2} = \cfrac{\pi}{\cfrac{7}{6}\pi} = \cfrac{6}{7}$

A particle moves according to the law

$x = r\cos \dfrac {\pi t}{2}$ . The distance covered by it in the time interval between

$t = 0$ and

$t = 3s$ is

0%
$r$
0%
$2r$
0%
$3r$
0%
$4r$

Explanation

Position of particle

$x = r\cos\dfrac{\pi t}{2}$

$t = 0\ s$ ,

$x = r\cos0 = r$

So, at

$t=0 \ s$ , particle is at extreme position of SHM.

Time period of oscillation

$T = \dfrac{2\pi}{w}= \dfrac{2\pi}{\pi/2} = 4 \ s$

In one time period, distance moved by particle

$= 4r$

So, in

$3 \ s$ distance moved

$= \dfrac{3}{4}\times 4r = 3r$

Two particle executing SHM of same frequency meet at

$x=+A/2$ , while moving in opposite direction. Phase difference between the particles is (

$A \rightarrow$ Amplitude of both)

0%
$\cfrac{\pi}{6}$
0%
$\cfrac{\pi}{3}$
0%
$\cfrac{5\pi}{6}$
0%
$\dfrac{-2\pi}{3}$

Explanation

For particle 1

$x_1 = a.\sin (wt)$ ...(1)

For particle 2

$x_2 = a.\sin (wt+y)$ ...(2)

For instance,

$x_1=x_2=\dfrac {a}{2}$

From (1)

$\dfrac {a}{2} = a.\sin (wt)$

Solving we get,

$wt=\dfrac{\pi}{6}$ or

$wt=\dfrac{5\pi}{6}$

Putting these value in (2),

$\sin \dfrac{\pi}{6} = a.\sin \left ( \dfrac{5\pi}{6} +y\right ) = \dfrac {a}{2}$

Solving this,

$y=-\dfrac{2\pi}{3}$

Two wires made of the same material, one thick and the other thin, are connected to form a composite wire. The composite wire is subjected to some tension. A wave travelling along the wire crosses the junction point. The characteristic that doesn't undergoes a change at the junction point is

0%
Frequency only
0%
Speed of propagation only
0%
Wavelength only
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The speed of propagation as well as the wavelength

Explanation

In case of composite wire:

1) frequency remains the same

2)wavelength change(decrease into thick wire because thick wire is a denser medium) because f=F

$\cfrac{V_1}{\lambda_1} = \cfrac{V_2}{\lambda_2}$

Also speed is decreased.

The function

$sin \omega t - cos \omega t$ represents

0%
a simple harmonic motion with a period $\frac{\pi}{\omega}$ .
0%
a simple harmonic motion with a period $\frac{2\pi}{\omega}$ .
0%
a periodic, but not simple harmonic motion with a period $\frac{\pi}{\omega}$ .
0%
a periodic, but not simple harmonic motion with a period $\frac 2{\pi}{\omega}$ .

Explanation

$x=\sin \omega t-\cos \omega t$

$\Longrightarrow x=2\cos \omega t=2\sin (\omega t+\cfrac{\pi}{2})\quad \quad \quad \quad [\sin C-\sin D=2\cos (\cfrac{C+D}{2})\sin (\cfrac{C-D}{2})]$

On comparing this with equation of

$SHM:$

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

$\Longrightarrow A=2\quad\quad \omega \text{(Angular frequency)}$

$\Longrightarrow T=\cfrac{2\pi}{\omega}$

A vibratory motion is represented by

$x = 2A cos \omega t + A cos [\omega t + \frac{\pi}{2}] + A cos (\omega t + \pi) + \frac{A}{2} cos [\omega t + \frac{3 \pi}{2}]$ .
The resultant amplitude of the motion is

0%
$\dfrac{9A}{2}$
0%
$\dfrac{\sqrt5A}{2}$
0%
$\dfrac{5}{2}A$
0%
$2A$

Explanation

We can solve this by a vector, See the figure I have given.

Resultant, see the figure ii.

$A_{net}=\sqrt{\left(2A-A\right)^2+{\left(A-\cfrac{A}{2}\right)^2}}=\sqrt{A^2+\cfrac{A^2}{4}}=\cfrac{\sqrt{5}A}{2}$

1152198_938029_ans_2faebcf9e09b48109ddd2f842fe04f4e.png

How long after the beginning of motion is the displacement of a harmonically oscillating particle equal to one half its amplitude if the period is

$24\ s$ and particle starts from rest

0%
$12\ s$
0%
$2\ s$
0%
$4\ s$
0%
$6\ s$

Explanation

Let's draw the phasor diagram

$\cos\theta=\cfrac{\cfrac{A}{2}}{A}=\cfrac{1}{2}\\ \theta 60°$

On rotation of

$960°$ it takes

$24$ sec the for

$60°$ rotation it will take

$\cfrac{24}{6}=4sec$

1151436_870051_ans_152618d9fbb8410c9d133fc09c33c353.png

A string of length 'l' is fixed at both ends. It is vibrating in its 3^rd overtone with maximum amplitude 'a'.The amplitude at a distance l/3 from one end is?

0%
$14a$
0%
$15$ a
0%
$10$ a
0%
None of these

Explanation

$y=(a\sin \omega t)\cos kx\\y=(a\sin \omega t)\cos(\cfrac{2\pi}{l/2})x\\At \quad x=l/2\\y=(a\sin\omega t)\cos(\cfrac{2\pi}{l}\times2\times\cfrac{l}{3})\\y=(a\sin\omega t)\cos(\cfrac{4\pi}{3})$

In this question, without calculating we can answer that none of these are correct.

Because given that maximum amplitude

$=a$ then how can at any place amplitude will be greater than

$'a'$ . It is not possible.

1188331_870975_ans_5f4007cc0c9743ab84c77c8de586a390.png

In a certain oscillatory system, the amplitude of motion is

$5$ m and the time period is

$4$ s. The time taken by the particle for passing between points which are at distance of

$4$ m and

$2$ m from the centre and on the same side of it will be

0%
$0.30$ s
0%
$0.32$
0%
$0.33$
0%
$0.35$

Explanation

Let the motion be represented by

$x=A\cos (wt)$

Given amplitude

$(A)=5$

$w=\dfrac{2\pi}{7}$ with periodic time

$T=45$

$\therefore w=\dfrac{2\pi}{4}=\dfrac{\pi}{2}$

Also

$x_{1}=4\ cm$

$x_{2}=2\ m$

Let

$t=t_{4}$ for displacement of

$4\ m$

$\therefore 4=5\cos \left(\dfrac{\pi}{2}\times t_{4}\right)$

$\therefore 0.8=\cos \left(\dfrac{\pi}{2}t_{4}\right)$

$\therefore t_{4}=\dfrac{36.83}{90}=$

For displacement of

$2\ m$ let

$t=t_{2}$

$\therefore 2=5\cos\left(\dfrac{\pi}{2}t_{2}\right)$

$\therefore 0.4\cos\left(\dfrac{\pi t_{2}}{2}\right)$

$\therefore t_{2}=\dfrac{66.5}{90}=$

$\therefore$ Time to travel from

$4\ m$ to

$2\ m$ is

$=t_{4}-t_{2}$

$=\dfrac{29.67}{90}=0.3248$ seconds

$\approx 0.33$ seconds

The equation of a wave is given by

$y= 10$ sin

$\left(\dfrac{2\pi}{45} t + \alpha \right)$ .
If the displacement is

$5 cm$ at

$t=0$ , the the total phase at

$t=7.5 s$ is

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

Explanation

The given equation of wave is

$y= 10$ sin

$\left(\dfrac{2\pi}{45} t + \alpha \right)$

At

$t= 0, y= 5 cm$

$\therefore 5= 10$ sin

$\alpha$

$\dfrac{1}{2}=$ sin

$\alpha$ or sin

$\left(\dfrac{\pi}{6}\right)=$ sin

$\alpha$

$\alpha = \dfrac{\pi}{6}$ ......(i)

hence, the total phase at

$t$ =

$7.5 s=\left(\dfrac{15}{2}s\right)$ is

$\phi=\dfrac{2\pi}{45} \times \dfrac{15}{2} + \alpha =\dfrac{\pi}{3} + \dfrac{\pi}{6}$ (Using (i))

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

A particle executes SHM with time period

$T$ and amplitude

$A$ . The maximum possible average velocity in time

${T}/{4}$ is:

0%
$\cfrac{2A}{T}$
0%
$\cfrac{4A}{T}$
0%
$\cfrac{8A}{T}$
0%
$\cfrac{4\sqrt {2}A}{T}$

Explanation

The maximum possible average velocity

$= \dfrac{\text{maximum displacement}}{\text{time}}$

i.e. Vavg

$= \dfrac{dmax}{T/4}$

here

$T \to 2\pi$

$\dfrac{T}{4} \to \dfrac{\pi}{2}$

$\therefore \theta = \dfrac{\pi}{4}$

$\therefore dmax = PQ = 2A\sin \theta$

$= 2a\sin \dfrac{\pi}{4}$

$\therefore dmax = \dfrac{2A}{\sqrt{2}} = \sqrt{2}A$

$\therefore Vavg = \dfrac{\sqrt{2}A}{T/4} = \dfrac{4\sqrt{2}A}{T}$

1456082_808769_ans_890f7a5a8ca54ee6b7ccea4c9878da7f.PNG

Two wires made of the same material, one thick and other thin, are connected to form a composite wire. The composite wire is subjected to some tension. A wave travelling along the wire crosses the junction point. The characteristic that undergoes a change at the junction point is:

0%
Frequency only
0%
Speed of propagation only
0%
Wavelength only
0%
The speed of propagation as well as the wavelength

Explanation

In case of composite wire

1) frequency remains same

2) wavelength change (decrease into thick wire because thick wire is a denser medium) because f=F

$\cfrac{v_1}{\lambda_1} = \cfrac{v_2}{\lambda_2}$

Also speed decreases.

Equation of plane progressive wave is given by

$y = 0.6 sin 2\pi\left(t - \dfrac{x}{2}\right)$ . On reflection from a denser medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is

0%
$y = 0.6 sin 2\pi\left(t + \dfrac{x}{2}\right)$
0%
$y =- 0.4 sin 2\pi\left(t + \dfrac{x}{2}\right)$
0%
$y = 0.4 sin 2\pi\left(t + \dfrac{x}{2}\right)$
0%
$y =- 0.4 sin 2\pi\left(t - \dfrac{x}{2}\right)$

Explanation

Amplitude of reflected wave

$= \dfrac{2}{3} \times 0.6 = 0.4$
On reflection from a denser medium there is phase change of

$\pi$ .

$\therefore$ Equation of reflected wave is

$y = 0.4 sin 2\pi\left(t + \dfrac{x}{2} + \pi\right) = - 0.4 sin 2\pi\left(t + \dfrac{x}{2}\right)$

Spherical wavefronts, emanating from a point source, Strike a plane reflecting surface. What will happen to these wave fronts, immediately after reflection ?

0%
They will remain spherical with the same curvature, both in magnitude and sign.
0%
They will become plane wave fronts.
0%
They will remain spherical, with the same curvature, but sign of curvature reversed.
0%
They will remain spherical, but with different curvature, both in magnitude and sign.

Assertion - On reflection from a rigid boundary there takes place a complete reversal of phase.
Reason - On reflection from a denser medium, both the particle velocity and wave velocity are reversed in sign.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

The reflection from a rigid boundary is same as the reflection from a medium of high density or a denser medium. When the reflection takes place, there is a reversal in the sign of the wave velocity as well as the particle velocity i.e. the complete reversal of the phase.

Thus, option

$A$ is correct.

Which of the following wave function does represent a travelling wave?

0%
$y=$ tan $(x - vt)^2$
0%
$y=$ log $(x- vt)$
0%
$y=\dfrac{1}{x+vt}$
0%
None of these

there are

$26$ tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives

$3$ beats with the next. The first one is octave of the last. What is the frequency of

$18^{th}$ tuning fork?

0%
$100 Hz$
0%
$99 Hz$
0%
$96 Hz$
0%
$103 Hz$

Explanation

Let the frequency of first tuning fork is

$\upsilon$ . The frequencies of other tuning forks are

$(\upsilon - 3), (\upsilon - 2 \times3), ......,(\upsilon - 17 \times 3), .....,(\upsilon - 25 \times 3)$ .
As per given condition,

$\upsilon = 2(\upsilon - 25 \times 3)$ or

$\upsilon = 2\upsilon - 25 \times 6$
or

$\upsilon = 25 \times 6 = 150 Hz$
The frequency of the

$18^{th}$ tuning fork

$\upsilon - 17 \times = 150 - 51 = 99 Hz$

A particle executes SHM of period

$12$ s. After two seconds, it passes through the centre of oscillation, the velocity is found to be

$3.142 cm s^{-1}$ . The amplitude of oscillation is:

0%
$6$ cm
0%
$3$ cm
0%
$24$ cm
0%
$12$ cm

Explanation

Let

$y = A sin \omega t$

Then

$v = \dfrac{dy}{dt} = \omega A cos \omega t = \dfrac{2 \pi}{T} A cos \dfrac{2 \pi}{T}t$

$\therefore 3.142 = \dfrac{2 \times 3.142}{12} \times A cos \dfrac{2 \pi}{12} \times 2$

$\therefore A = 12cm$

The amplitude of an electromagnetic wave in vacuum is doubled with no other changes made to the wave. As a result of this doubling of amplitude, which of the following statement is correct?

0%
The speed of the wave propagation changes only
0%
The frequency of the wave changes only
0%
The wavelength of the waves changes only
0%
None of these

Explanation

Velocity of EM wave is given by,

$v =\dfrac{1}{\sqrt{\epsilon_0 \mu_0}}$

This speed of the wave is the same as the speed of light in vacuum.

$v=3*10^8m/s$

This velocity of the electromagnetic wave is independent of the amplitude of the wave and also independent of frequency.

The displacement of a particle executing simple harmonic motion is given by

$x = 3 sin [2 \pi t + \frac{\pi}{4}]$ where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is:

0%
$3m, 2 \pi m s^{-1}$
0%
$3m, 4 \pi m s^{-1}$
0%
$3m, 6 \pi m s^{-1}$
0%
$3m, 8 \pi m s^{-1}$

Explanation

The given equation of SHM is

$x = 3sin[2 \pi t + \dfrac{\pi}{4}]$

Compare the given equation with standard equation of SHM

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

we get,

$A = 3m, \omega = 2 \pi s^{-1}$

$\therefore Maximum \,\, speed, v_{max} = A \omega = 3 m \times 2 \pi s^{-1} = 6 \pi m s^{-1}$

Two particles execute simple harmonic motions of same amplitude and frequency along the same straight line. They cross one another when going opposite directions. The phase difference between them when their displacements are one half of their amplitudes is then

0%
$60^0$
0%
$30^0$
0%
$120^0$
0%
$150^0$

Explanation

The equation for SHM is,

$y = A sin(\omega + \phi)$
As the displacement is half of the amplitudes

$(y = \dfrac{A}{2})$ , or

$\dfrac{A}{2} = A sin (\omega t + \phi) or sin (\omega t + \phi) = \dfrac{1}{2}$

$\therefore \omega t + \phi = 30^0 or 150^0$ .
Since the two particles are going in opposite directions, the phase of one is

$30^0$ and that of the other

$150^{0}$ . Hence the phase difference between the two particles =

$150^0 - 30^0 = 120^0$ .

The equation of the stationary wave is

$y = 2A sin(\dfrac{2 \pi c t}{\lambda}) cos (\dfrac{2 \pi x}{\lambda})$
Which of the following statements is wrong?

0%
The unit of ct is same as that of $\lambda$ .
0%
The unit of x. is same as that of $\lambda$
0%
The unit of ct is same as that of $x$
0%
The unit of ct, $\lambda$ and $x$ are same

Explanation

$y=2A\sin\left( \dfrac { 2\Pi t }{ \lambda } \right) \cos\left( \dfrac { 2\Pi x }{ \lambda } \right)$

Arguments of

$sin$ and

$cos$ are unit less.

$\Rightarrow \dfrac { Ct }{ \lambda } ,\dfrac { x }{ \lambda }$ have dimension

${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$ .

$\Rightarrow$

$Ct,x$ and

$\lambda$ have same dimensions.

In the standing waves that form as a result of reflection of waves from an obstacle, the ratio of the amplitude at antinode to the amplitude at a node is s. The fraction of the energy that passes past the obstacle is:

0%
$\dfrac{4}{s^2}$
0%
$\dfrac{s + 1}{s - 1}$
0%
$(\dfrac{s + 1}{s - 1})^2$
0%
non of the above.

Explanation

Let

$A_0$ be amplitux of original wave

$A_r$ be amplitux of reflected wave

$s/a$

$\dfrac{A_0+A_r}{A_0-A_r}=S$

Applying comp. and dive.

$\dfrac{A_0+A_r+A_0-A_r}{A_0+A_r-A_0+A_r}=\dfrac{s+1}{s-1}$

$\Rightarrow A_r=\dfrac{s-1}{s+1}A_0$

Now

$4\propto A^2$

$\therefore$ Req ratio

$\dfrac{A_0^2-A_r^2}{A_0^2}=1-\left(\dfrac{s-1}{s+1}\right)^2$

$=\dfrac{2s}{(s+1)^2}$

Remark: Non of option is correct

The maximum velocity of a body undergoing S.H.M. is

$0.2 m/s$ and its acceleration at

$0.1 m$ from the mean position is

$0.4 m/s^{-2}$ The amplitude of the S.H.M. is:

0%
$0.25 m$
0%
$0.3 m$
0%
$0.1 m$
0%
$1.05 m$

Explanation

${V}_{max}= \omega A= 0.2 m/s$

$a--{ \omega }^{ 2 }x=0.4 m/{s}^{2}$

${ \omega }^{ 2 }=4$

$\Rightarrow$

$(\omega =2)$

$\omega A=0.2$

$2 \times A=0.2$

$(A=0.1m)$

If the particle was to start at the extreme position i.e x = +A, then what will be the equation of SHM:

0%
$A cos (\omega t)$
0%
$A sin (\omega t)$
0%
$A cot (\omega t)$
0%
$A tan (\omega t)$

Explanation

The equation of SHM, when the particle started at the mean position at t = 0 is given by

$x=A sin (\omega t + \phi); \phi$ is the phase constant.

At t =0, if x=A, then, we have

$sin \phi =1$ or

$\phi=\pi/2$

Thus, the equation of SHM will be

$x=A sin(\omega t + \pi/2)=A cos (\omega t)$

The correct option is (a)

At a particular position the velocity of a particle in SHM with amplitude a is $\dfrac{{\sqrt 3 }}{2}$ that at its mean position. In this position, its displacement is:

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

Explanation

Let the equation of SHM is

$x=a sin(\omega t)$ ,

$\therefore v=a\omega cos(\omega t)$

$\Rightarrow$ Velocity at mean position

$v= a\omega$ ,

As given in question,

$\dfrac{\sqrt 3}{2}a\omega= a\omega cos (\omega t)$

$\Rightarrow cos(\omega t) = \dfrac{\sqrt 3}{2}$

$\Rightarrow \omega t= \dfrac{\pi}{6}$

So Displacement

$x= a sin(\dfrac{\pi}{6})=\dfrac{a}{2}$

A particle executes SHM given by an equation

$x=A sin (\omega t + \pi/6)$ . Where was the particle at t = 0

0%
A
0%
Mean position
0%
A/2
0%
A/4

Tapping gently on the surface of water produces concentric ripples. This is due to

0%
circular interference of waves
0%
linear interference of waves
0%
energy transfer radially
0%
vibration of nearby particles

A plane wave of wavelength

$\lambda$ is incident at an angle

$\theta$ on a plane mirror. Maximum intensity will be observed at

$P$ , when

0%
$\cos \theta = 3\lambda / 2d$
0%
$\sec \theta - \cos \theta = 3\lambda / 4d$
0%
$\cos \theta = \lambda / 4d$
0%
$\sec \theta - \cos \theta = \lambda / 4d$

Explanation

for max intensity

$CO+OP=$ multiple of

$\cfrac { \lambda }{ 2 }$

from fig.

$\cfrac { dcos{ 2\theta } }{ cos{ \theta } } +\cfrac { d }{ cos{ \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cfrac { d(1+cos{ 2\theta }) }{ \cos { \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cfrac { d2\cos { ^{ 2 }\theta } }{ \cos { \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cos { \theta } =\cfrac { \lambda }{ 4d }$

1128237_1010767_ans_c3fae067bca8456799c2a98584c78d2b.png

A particle moves according to

$x= a cos \Big(\frac{\pi}{2}t\Big)$ . The distance covered by it in time interval between t=0 to t=2 s is

0%
2a
0%
3a
0%
4a
0%
a

Explanation

$\dfrac{\pi}{2}=\frac{2\pi}{T} \Rightarrow T= 4$ sec

$S_{0-2}= 2a$

Given that

$y = A \sin \left[\left(\dfrac{2 \pi}{\lambda}\right)(ct - x)\right]$ , where

$y$ and

$x$ are measured in the unit of length. Which of the following statements is true?

0%
The unit of $\lambda$ is same that of $x$ and $A$ .
0%
The unit of $\lambda$ is same that of $x$ but may not be same as that of $A$ .
0%
The unit of $c$ is same as that of $2 \pi/ \lambda$
0%
The unit of $(ct - x)$ is same as that of $2 \pi/ \lambda$

Explanation

$y=A\sin\left( \dfrac { 2\Pi }{ \lambda } \left( ct-x \right) \right)$

$\dfrac { ct-x }{ \lambda }$ is written

$\Rightarrow$

$x$ and

$\lambda$ have same unit.

$\sin\left( \theta \right)$ is unitless

$\Rightarrow$

$y$ and

$A$ have same unit.

Which of these does not represent a wave equation

0%
$y=A sin (\omega t- kx)$
0%
$y=A sin (\omega t + kx)$
0%
$y=A cos (\omega t- kx)$
0%
$y=A (\omega t- kx)$

A travelling wave is represented by the equation

$y=20 sin 2(x-4t)$ . If y is measured in cms, what is the amplitude of the wave

0%
20 mms
0%
20 cm
0%
20 m
0%
2 cm

A travelling wave has a velocity of 400 m/s and has a wavelength of 0.5 m. What is the phase difference between two points in the wave that are 1.25 milli secs apart

0%
$2 \pi$
0%
$2 \pi/3$
0%
$2 \pi/5$
0%
$\pi/6$

Explanation

The frequency of the wave is

$f = v/\lambda=400/0.5=800 Hz$

Time period = 1/800 = 1.25 ms

Thus, both the points are one time period apart. Hence their phase difference will be zero or

$2 \pi$

The correct option is (a)

The displacement of a particle varies according to the relation

$x= 4 (cos \pi t + sin \pi t)$ . The amplitude of the particle is

0%
$-4$
0%
$4$
0%
$4\sqrt2$
0%
$8$

Explanation

Given that,

The displacement is

$x=4\left( \cos \pi t+\sin \pi t \right)$

$x=4\cos \pi t+4\sin \pi t$

Let,

$R\cos \phi =4....(I)$

$R\sin \phi =4.....(II)$

Now,

$x=R\sin \phi \cos Pt+R\cos \phi \sin \pi t$

$x=Rsin\left( \pi t+\phi \right)$

So, R is the amplitude

Now, squaring and adding equation (I) and (II)

$R=\sqrt{{{4}^{2}}+{{4}^{2}}}$

$R=4\sqrt{2}$

Hence, the amplitude is $4\sqrt{2}$

A progressive wave of wavelength 5 cm moves along +X axis. What is the phase difference between two points on the wave separated by a distance of 3 cm at any instant

0%
$3 \pi/5$
0%
$6 \pi/5$
0%
$2 \pi/5$
0%
$7 \pi/5$

Explanation

Phase difference =

$(2\pi/5)$ path difference

$\implies$ phase difference =

$6 \pi/5$

The correct option is (b)

The equation of a travelling wave is given by

$3sin(\pi/2)(25t-x)$ where x and y are in meters and t is in seconds. What is the ratio of maximum amplitude to its wavelength

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

Explanation

$y = 3 sin (\pi/2) (50t - x)$

The wavelength is

$\lambda=(2 \pi/k)=2 \pi/(\pi/2)=4$

The maximum amplitude is 3. Hence ratio = 3/4

The correct answer is (c)

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude

$A_1$ . When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude

$A_2$ . The ratio of

$\left( \frac{A_1}{A_2}\right)$ is:

0%
$\dfrac{M}{m}$
0%
$\left(\dfrac{M}{M+m}\right)^{1/2}$
0%
$(M+m)\sqrt{\dfrac{k}{M+m}A_2}$
0%
$\dfrac{M}{M+m}$

Explanation

$COM \Rightarrow M \sqrt{\dfrac{k}{M}A_1}= (M+m)\sqrt{\dfrac{k}{M+m}A_2}$

A magnifier lens focusses a beam of light on a sheet of paper for 5 secs and the paper starts burning. The process described here is

0%
Concentrated light produces heat
0%
Light energy is converted in to heat energy
0%
Photons heating the paper
0%
Electrons heating the paper

Phase difference between a compression and its successful rarefaction is

$2 \pi$ radians

0%
True
0%
False

Explanation

Phase difference between a crest and its successive trough is

$\pi$ radians

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:

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

Explanation

$x1=Asin \big( \omega t + \phi 1\big)$

$x2=Asin \big( \omega t + \phi 2\big)$

$x1-x2=A \big(2sin \big( \omega t+ \frac{ \phi1 + \phi2 }{2} \big)sin \big( \frac{ \phi1 - \phi2}{2} \big) \big)$

$A=2Asin \frac{ \phi1- \phi2 }{2}$

$\frac{ \phi1- \phi2 }{2}= \frac{ \Pi }{6}$

$\phi1=\frac{ \Pi }{3}$

Two waves of same amplitude and frequency with a phase difference of

$\pi/2$ and travelling in same direction are superimposed each other, the maximum amplitude of the resultant wave is

0%
2A
0%
A
0%
A $\sqrt{2}$
0%
A/2

Explanation

The resultant of two waves is given by

$R^2=A^2+B^2+2AB cos \theta; \theta$ is the phase difference and A and B are the amplitudes of the individual waves.

Since both the waves are identical, their amplitudes are equal. The phase difference between the wave is

$\pi/2$ . Substituting, we get,

Thus,

$R = A \sqrt(2)$

The correct option is (c)

CBSE Questions for Class 11 Engineering Physics Waves Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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