MCQ Questions for Class 11 Engineering Physics Waves Quiz 7

If two waves of same frequency and amplitude respectively on superposition produce a resultant disturbance of the same amplitude, the wave differ in phase by :

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

Explanation

Resultant amplitude due to superposition of two waves with phase difference

$\phi$ is given by

$A^2=A_1^2+A_2^2+2A_1A_2cos\phi$

It is given that

$A_1=A_2=A$

Thus

$A^2=A^2+A^2+2A^2cos\phi$

$\implies cos\phi=-\dfrac{1}{2}$

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

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%
$4\sqrt {2}$
0%
$8$

Explanation

Hint: Here, we have to apply trigonometric formulas

Solution:

Step1: Simplify the given displacement

$x = 4\left( {cos\pi t{\text{ }} + {\text{ }}sin\pi t} \right).....(1)$

Multiply & divide each term of R.H.S. of equation (1) by $\sqrt 2$ we get,

$\Rightarrow \dfrac{4}{{\sqrt 2 }} \times \sqrt 2 \left( {cos\pi t{\text{ }} + sin\pi t} \right)$

$\Rightarrow 4\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\cos \pi t + \dfrac{1}{{\sqrt 2 }}\sin \pi t} \right).....(2)$

Step2: Apply required trigonometric formula
we know that, $\dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}$ and $\dfrac{1}{{\sqrt 2 }} = \cos \dfrac{\pi }{4}$

Now, equation (2) becomes,

$\Rightarrow 4\sqrt 2 \left( {\cos \pi t\sin \dfrac{\pi }{4} + \sin \pi t\cos \dfrac{\pi }{4}} \right).....(3)$

Also, $\sin (a + b) = \sin a\cos b + \cos a\sin b$

Now, equation (3) becomes,

$x = 4\sqrt 2 \sin \left( {\pi t{\text{ }} + \dfrac{\pi }{4}} \right).....(4)$

Step3: Find Amplitude ‘A’

Comparing above equation (4) with,

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

We get, Amplitude, $A = 4\sqrt 2$

Hence, option (C) is correct.

A progressive wave of frequency

$500\ Hz$ is travelling with a speed of

$330\ m/s$ in air. The distance between the two points which have a phase difference of

$\displaystyle { 30 }^{ \circ }$ is:

0%
$0.11\ m$
0%
$0.055 \ m$
0%
$0.22\ m$
0%
$0.025\ m$

Explanation

Wavelength of the wave is

$\lambda =\dfrac { v }{ f } =\dfrac { 330 }{ 500 }\ m =\dfrac { 33 }{ 50 }\ m$

$k=\dfrac { 2\pi }{ \lambda } =\dfrac { 100\pi }{ 33 }$

Since, phase difference is given by

$k\Delta x$

$\Delta \phi=k\Delta x=\dfrac { 100\pi }{ 33 } \Delta x=\dfrac { \pi }{ 6 }$

$\Delta x=0.055\ m$

In a medium in which, a transverse progressive wave travelling, the phase difference between the points with a separation of

$1.25\ cm$ is

$\displaystyle { \pi }/{ 4 }$ . If the frequency of wave

$1000\ Hz$ , the velocity in the medium is :

0%
$\displaystyle { 10 }^{ 4 }\ { ms }^{ -1 }$
0%
$\displaystyle 125\ { ms }^{ -1 }$
0%
$\displaystyle 100\ { ms }^{ -1 }$
0%
$\displaystyle 10\ { ms }^{ -1 }$

Explanation

Let

$x_1$ and

$x_2$ be two points,
then, the phase difference between two points

$x_1 and \ x_2$ is

$\dfrac{2\pi}{\lambda}(x_2-x_1)$

$\implies \dfrac{2\pi}{\lambda}(1.25\times 10^{-2})=\dfrac{\pi}{4}$

$\implies \lambda=0.1m$

Thus, speed of wave in the medium =

$\lambda\nu=0.1\times 1000=100\ m/s$

Find out the change in wavelength of a

$700 Hz$ acoustic wave as it enters brass from warm air? Given Speed of Sound in warm air is

$350 {m}/{s}$ and through brass is

$3,500 {m}/{s}$ .

0%
It decreases by a factor of $20$
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It decreases by a factor of $10$
0%
It increases by a factor of $10$
0%
It increases by a factor of $20$
0%
The wavelength remains unchanged when a wave passes into a new medium

Explanation

Given :

$\nu = 700$

$Hz$

$v_a = 350 m/s$

$v_b = 3500 m/s$

Using

$v = \nu \lambda$

$\therefore$

$\lambda_a = \dfrac{v_a}{\nu} = \dfrac{350}{700} =0.5$

$m$

Similarly

$\lambda_b = \dfrac{v_b}{\nu} = \dfrac{3500}{700} =5$

$m$

Thus wavelength increases by a factor of

$10$

Five sinusoidal waves have the same frequency

$500Hz$ but their amplitudes are in the ratio

$2:\cfrac{1}{2}:\cfrac{1}{2}:1:1$ and their phase angles

$0,\cfrac{\pi}{6},\cfrac{\pi}{3},\cfrac{\pi}{2}$ and

$\pi$ respectively. The phase angle of resultant wave obtained by the superposition of these five waves is:

0%
${30}^{o}$
0%
${45}^{o}$
0%
${60}^{o}$
0%
${90}^{o}$

Explanation

The phasors of five waves can be represented as

$x_1=2\hat i$

$x_2=\sqrt 3/4 \hat i + 1/4 \hat j$

$x_3=1/4 \hat i + \sqrt 3 /4 \hat j$

$x_4=\hat j$

$x_5=-\hat i$

Resultant

$x=\sum_{i=1}^5 x_i$

$x=\cfrac{5+\sqrt 3}{4}\hat i + \cfrac{5+\sqrt 3}{4}\hat j$

Phase angle of resultant,

$\theta =tan^{-1}\dfrac{(5+\sqrt 3)/4}{(5+\sqrt 3)/4}$

$\Rightarrow \theta = 45^o$

If n represents the order of a half period zone the area of this zone is approximately proportional to

$n^{m}$ where m is equal to

0%
zero
0%
half
0%
one
0%
two

Explanation

Area of half period zone is independent of order of zone. Therefore, m is equal to zero in

$n^m$ .

Two simple harmonic motions are given by:

${ x }_{ 1 }=a\sin { \omega t } +a\cos { \omega t }$

${ x }_{ 2 }=a\sin { \omega t } +\cfrac { a }{ \sqrt { 3 } } \cos { \omega t }$
The ratio of the amplitudes of first and second motion and the phase difference between them are respectively:

0%
$\sqrt { \cfrac { 3 }{ 2 } }$ and $\cfrac { \pi }{ 12 }$
0%
$\cfrac { \sqrt { 3 } }{ 2 }$ and $\cfrac { \pi }{ 12 }$
0%
$\cfrac { 2 }{ \sqrt { 3 } }$ and $\cfrac { \pi }{ 12 }$
0%
$\sqrt { \cfrac { 3 }{ 2 } }$ and $\cfrac { \pi }{ 6 }$

Explanation

$\Rightarrow { x }_{ 1 }=a\sin { \omega t } +a\cos { \omega t }$

$=$

$a\sqrt { 2 } \sin { (\omega t } +45)$

$\Rightarrow { x }_{ 2 }=a\sin { \omega t } +\cfrac { a }{ \sqrt { 3 } } \cos { \omega t } = \dfrac { 2a }{ \sqrt { 3 } } \sin ({ \omega t }+30)$

Using these equations above phasor diagrams are drawn.

Ratio of amplitude

$=\cfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\cfrac { \sqrt { 3 } }{ \sqrt { 2 } }$

Phase difference

$=\cfrac { \pi }{ 4 } -\cfrac { \pi }{ 6 } =\cfrac { \pi }{ 12 }$

A pulse of a wave travelling on a string towards the fixed rigid end as shown in the figure above. The pulse is:

0%
Reflected and transmitted
0%
Reflected and refracted
0%
Reflected and reduced
0%
Reflected and magnified
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Reflected and inverted

State whether the given statement is True or False :

If the distance between two successive troughs in a transverse wave is

$8 cm$ , then the amplitude of that wave is

$4 cm$ .

0%
True
0%
False

Explanation

$False$

The distance between two successive troughs or crests in a transverse wave is known as the wavelength of the wave.

The maximum displacement of the particles of the medium from their mean positions is known as the amplitude.

Usually, wavelength and amplitude are independant of each other.

The information given in the question is inadequate for calculating the amplitude.

A particle in SHM is described by the displacement function:

$x(t)= A \ cos (\omega t+ \phi).$ If the initial (t=0) position of the particle is 1 cm and its initial velocity is

$\pi$ cm/sec , what is the amplitude?

0%
$\sqrt 2 cm$
0%
$2 cm$
0%
$\sqrt 7 cm$
0%
$12 cm$

Explanation

$x(t)=Acos(\omega t+\phi)$

$\implies v(t)=\dfrac{dx(t)}{dt}$

$=-\omega Asin(\omega t+\phi)$

Initial position=

$x(t=0)=Acos\phi=1cm$

Initial velocity=

$v(t=0)=-A\omega sin\phi=\omega cm/s$

Hence,

$A=\sqrt{2}cm$

A particle in SHM is described by the displacement function:

$x(t)= A \ cos (\omega t+ \phi).$ If the initial

$(t=0)$ position of the particle is

$1$ cm and its initial velocity is

$\omega$ cm/sec , what is the initial phase angle ?

0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{7 \pi}{4}$
0%
$\dfrac{7 \pi}{2}$
0%
$\dfrac{\pi}{4}$

Explanation

$x(t)=Acos(\omega t+\phi)$

$\implies v(t)=\dfrac{dx(t)}{dt}$

$=-\omega Asin(\omega t+\phi)$

Initial position=

$x(t=0)=Acos\phi=1cm$

Initial velocity=

$v(t=0)=-A\omega sin\phi=\omega cm/s$

Hence,

$\phi=tan^{-1}(-1)=\dfrac{7\pi}{4}$

The wave carries ______________

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Power
0%
Energy
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Displacement
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Work

The frequency of a sound wave is 200 Hz and its wavelength is 150 cm. What is the distance travelled by the sound wave in the time taken to produce 150 waves?

0%
110 cm
0%
225 cm
0%
112.5 cm
0%
336.5 cm

Explanation

As frequency is

$200\,Hz$ , that means

$200$ waves are produced on one second.

So, time taken to produce

$150$ waves

$=\frac{150}{200}=0.75\,s$

Velocity of wave = Frequency

$\times$ Wavelength

$=200\times 1580=30000$ cmps

$=300$ meters per second

So, Distance

$=300\times 0.75=225\,cmeters$

Two travelling waves of equal frequency, with amplitude

$4 cm$ and

$6 cm$ respectively, superimpose in a single medium. Find out the range of the amplitude,

$A$ , of the resultant wave?

0%
$2 cm \leq A \leq 10 cm$
0%
$A = 5 cm$
0%
$A = 10 cm$
0%
$10 cm \leq A \leq 12 cm$
0%
$12 cm \leq A \leq 24 cm$

Explanation

Given :

$A_1 = 4$ cm,

$A_2 =6 cm$

Maximum superimposition: The two waves interfere constructively.

Resultant amplitude

$A' = A_1 + A_2 = 4 + 6 =10 cm$

Minimum superimposition: The two waves interfere destructively.

Resultant amplitude

$A'' = A_2 - A_1 = 6 - 4 =2 cm$

Thus range of amplitude is

$2 cm \leq A \leq 10 cm$

Two simple harmonic motions are given by

$x_1 = a \sin \omega t + a \cos \omega t$ and

$x_2 = a \sin \omega t + \dfrac{a}{\sqrt{3}}\cos \omega t$
The ratio of the amplitudes of first and second motion and the phase difference between them are respectively

0%
$\sqrt{\dfrac{3}{2}}$ and $\dfrac{\pi}{12}$
0%
${\dfrac{\sqrt{3}}{2}}$ and $\dfrac{\pi}{12}$
0%
${\dfrac{3}{\sqrt{2}}}$ and $\dfrac{\pi}{12}$
0%
$\sqrt{\dfrac{3}{2}}$ and $\dfrac{\pi}{6}$

Explanation

First SHM is given as

$x_1 = a\sin wt + a \cos wt = a [\sin wt + \cos wt]$

$\therefore$

$x_1 = \sqrt{2}a [\dfrac{1}{\sqrt{2}}\sin wt + \dfrac{1}{\sqrt{2}} \cos wt]$

$x_1 = \sqrt{2}a [\cos(\pi /4)\sin wt + \sin{\pi/4} \cos wt]$

$\implies$

$x_1 = \sqrt{2} a \sin (wt + \pi/4)$

Second SHM is given as

$x_2 = a\sin wt + \dfrac{a}{\sqrt{3}} \cos wt = a [\sin wt + \dfrac{1}{\sqrt{3}}\cos wt]$

$\therefore$

$x_2 = \dfrac{2}{\sqrt{3}}a [\dfrac{\sqrt{3}}{{2}}\sin wt + \dfrac{\sqrt{3}}{2\sqrt{3}} \cos wt]$

$x_2 = \dfrac{2}{\sqrt{3}}a [\dfrac{\sqrt{3}}{{2}}\sin wt + \dfrac{1}{2} \cos wt]$

$x_2 = \dfrac{2}{\sqrt{3}}a [\cos(\pi /6)\sin wt + \sin{\pi/6} \cos wt]$

$\implies$

$x_2 = \dfrac{2}{\sqrt{3}} a \sin (wt + \pi/6)$

Thus ratio of magnitude

$= \dfrac{\sqrt{2}a}{2a/\sqrt{3}} = \sqrt{\dfrac{3}{2}}$

Also phase difference

$\Delta \phi = \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{\pi}{12}$

An excerpt from a book by Einstein and Infeld gives the following remarks concerning wave phenomena:
"A bit of gossip starting in Washington reaches New York [by word of mouth] very quickly, even though not a single individual who takes part in spreading it travels between these two cities. There are two quite different motions involved, that of the rumor, Washington to New York, and that of the persons who spread the rumor."

Identify a correct inference from the above text.

0%
The particles of the medium perform motion from one place to another.
0%
The particles of the medium perform random motion to constitute wave motion.
0%
The particles constituting the medium perform only small vibrations, but the whole motion is that of a progressive wave.
0%
None of these.

The S.I unit of wavelength is ______________

0%
$cm$
0%
$dm$
0%
$m$
0%
$km$

During the propagation of wave motion,

0%
there is transfer of energy from one particle to another without any actual transfer of the particles of the medium.
0%
there is transfer of energy from one particle to another with transfer of the particles of the medium.
0%
there is no transfer of energy from one particle to another.
0%
None of these.

Identify the correct statement:

0%
The power of ripples in water is much greater than the power of sound waves generated by a single human voice.
0%
The power of ocean waves during a storm is much greater than the power of sound waves generated by a single human voice.
0%
The power of sound waves generated by a single human voice is much greater than the power of ocean waves during a storm.
0%
All statements are wrong.

Which of the following is a Longitudinal wave? _____________

0%
Radio wave
0%
X-rays
0%
Light waves
0%
Sound waves

As waves propagate through a medium, they transport energy. We can easily demonstrate this by

0%
hanging an object on a stretched string and then sending a pulse down the string
0%
throwing a stone into a pond
0%
compressing a spring and releasing it.
0%
None of these.

Three waves of amplitudes

$12\mu m$ ,

$4\mu m$ and

$9\mu m$ but of same frequency arrive at a point in a medium with a successive phase diffeerence of

$\dfrac{\pi}{2}$ . Then the resultant amplitude is

0%
$4$
0%
$7$
0%
$5$
0%
$25$

Explanation

Since three waves have a successive phase difference of

$\pi/2$ .
So, first and third waves have phase difference of

$\pi$ .
Thus, net amplitude of first and third wave is

$A_1 = 12-9 = 3\mu m$
Amplitude of second wave

$A_2 = 4\mu m$
Resultant amplitude

$A_R = \sqrt{A_1^2 + A_2^2}$

$\therefore$

$A_R = \sqrt{3^2 + 4^2} = 5\mu m$

The displacement y of a wave traveling in the X-direction is given by

$y = 10^{-4}\sin({600t-2x+\dfrac{\pi}{3})}$ metres. where

$x$ is expressed in metres and

$t$ in

$seconds$ . The speed of the wave motion in

$ms^{-1}$ is:

0%
$200$
0%
$300$
0%
$600$
0%
$1200$

Explanation

$y={ 10 }^{ -4 }sin\left( 600t-2x+\dfrac { \pi }{ 3 } \right)$

Here,

$W=600/s$

$K=2/m$

Speed of wave

$V=W/K$

$=600/2$

$=300m/s$

$\therefore$ Option (B) is correct.

A traveling wave in a stretched string is described by the equation,

$y = A \sin(kx+t)$ . The maximum particle velocity is

0%
A
0%
$k$
0%
$Ak$
0%
none of the above

Explanation

$v=\dfrac{dy}{dt}=Acos(kx+t)$

Since the max value of

$cos$ is 1.

The maximum velocity is

$A$ .

A weight of 5 kg is required to produce the fundamental frequency of a sonometer wire. What weight is required to produce its octave?

0%
20
0%
32
0%
26
0%
none of the above

Explanation

In a sonometer wire,

$f \propto \sqrt{T}$

$f_{octave}=2f_{fundamental}$

$T_{octave}=4T_{fundamental}$

$M_{octave}g=4m_{fundamental}$

$M_{octave}=20kg$

A stationary wave is a multidimensional variant depending on

0%
time axis
0%
position axis
0%
time and position axis
0%
none of the above

Explanation

A stationary wave is a multidimensional variation which oscillates in time but whose peak amplitude profile dose not move in space.

Equ of stationary wave :

$y=A\sin(wt)\cos(Kr)$

It depends on both time and position axis.

$\therefore$ Option (C) is correct.

Two sound waves each of amplitude

$a$ and of frequencies

$500 Hz$ and

$512 Hz$ superpose. Then the resultant of the amplitude of the resultant wave at

$t=\dfrac{5}{48} s$ is

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

Explanation

When two sound waves of slightly different frequencies and equal amplitudes superimpose on each other, then their resultant amplitude is given by

$R= 2a\cos(2\pi\dfrac{n_1-n_2}{2}t)=2a\cos(2\pi\times\dfrac{512-500}{2}\times\dfrac{5}{48})=\dfrac{a}{\sqrt2}$

Note that here the angle is in radians.

Substitute and calculate.

If two waves of length

$50 cm$ and

$51 cm$ produced

$12$ beats

$per\ second$ , the velocity of sound is:

0%
$360$
0%
$340$
0%
$306$
0%
none of the above

Explanation

Given,

Wavelength,

${ \lambda }_{ 1 }=50cm=0.5m\\ { \lambda }_{ 2 }=51cm=0.51m$

Frequency of the beats, f=12

Now,

$f=\frac { V }{ { \lambda }_{ 1 } } -\frac { V }{ { \lambda }_{ 2 } } =V\left[ \frac { 1 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } \right] \\ 12=V\left[ \frac { 1 }{ 0.5 } -\frac { 1 }{ 0.51 } \right] \\ 12=V\left[ \frac { 0.01 }{ 0.5\times 0.51 } \right] \\ \left[ V=306m/s \right]$

Hence option (C) is correct

A stationary wave is represented by the equation,

$y = 3 \cos(x/8) \sin(15t)$ where

$x$ and

$y$ are in

$cm$ and

$t$ is in

$seconds$ . The distance between the consecutive nodes is (in

$cm$ ):

0%
$8$
0%
$12$
0%
$14$
0%
$16$

Explanation

Distance between two consecutive nodes=

$\dfrac { \lambda }{ 4 }$

$\dfrac { \lambda }{ 4 } =\dfrac { 2\pi }{ K } \times \dfrac { 1 }{ 4 }$

From,

$y=3\cos { \left( \dfrac { x }{ 8 } \right) } \sin { \left( 15t \right) } \\ K=\dfrac { 1 }{ 8 } \\ \therefore \dfrac { \lambda }{ 4 } =\dfrac { 2\pi }{ K } \times \dfrac { 1 }{ 4 } =\dfrac { 2\pi \times 8 }{ 4 } =4\pi$

Hence distance between the consecutive nodes is

$4\pi$ cm

Two points on a travelling wave having frequency

$500 Hz$ and velocity

$300{ m }/{ s }$ are

${ 60 }^{ o }$ out of phase, then the minimum distance between the two points is

0%
$0.2$
0%
$0.1$
0%
$0.5$
0%
$0.4$

Explanation

Wavelength of the wave is:

$\lambda = \dfrac{v}{f} = \dfrac{300}{500} = 0.6\ m$

$360^o$ phase difference corresponds to a distance of

$1$ wavelength or

$360^o$ phase difference corresponds to a distance of

$0.6\ m$

$60^o$ phase difference corresponds to a distance of

$0.6/6 = 0.1\ m$

Sum of two mechanical waves travelling in any direction (having the same frequency),

0%
can be summed up to give a stationary wave
0%
is added using the phasor diagram
0%
can have graphical representation with no troughs and elevations
0%
none of the above

Stethoscope of doctors for finding quality, strength and frequency of human heart beat is based on the principle of

0%
SONAR
0%
Reverberation
0%
Multiple reflection
0%
Echo

A progressive wave is represented by y = 12 sin (5t - 4x) cm. On this wave, how far away are the two points having phase difference of 90

$^o$ ?

0%
$\dfrac{\pi}{2} cm$
0%
$\dfrac{\pi}{4} cm$
0%
$\dfrac{\pi}{8} cm$
0%
$\dfrac{\pi}{16} cm$

Explanation

Given equation for the wave is

$y = 12 \textrm{sin}(5t-4x)$

Phase is given by

$5t-4x$ . At a given instant, let two points

$x_1$ and

$x_2$ differ in phase by

$90^o = \pi/2$ .

$\Rightarrow (5t-4x_1)-(5t-4x_2) = \pi/2$

$\Rightarrow 4|x_2-x_1| = \pi/2$

Thus, distance between the two points is

$\pi/8$

$\textrm{cm}$

On watching the graphical representation of different waves which of the following does not have energy carrying capacity

0%
Electromagnetic waves
0%
Stationary waves
0%
Unidirectional travelling string wave
0%
none of the above

The equation of a simple harmonic wave is given by

$y = 6 \sin{2\pi} \left(2t-0.1x\right)$ , 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%
${18}^{o}$
0%
${36}^{o}$
0%
${54}^{o}$
0%
${72}^{o}$

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

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

The amplitude of resulting wave dues to superposition of

$y_1 = A sin (t -kx)$ &

$y_2 = A sin (t -kx + \delta)$ is

0%
$2A cos(\delta/2)$
0%
$2A tan (\delta/2)$
0%
$A \cos (\delta /2)$
0%
none

The amplitude of resulting wave dues to superposition of

$y_1 = A sin (t -kx)$ &

$y_2 = A sin (t- kx + \delta)$ is

0%
$2A cos(\delta/2)$
0%
$2A tan (\delta/2)$
0%
$A cos(\delta/2)$
0%
none

The path difference between two waves

$y_{1} = a_{1}\sin \left (\omega t - \dfrac {2\pi x}{\lambda}\right )$ and

$y_{2} = a_{2}\cos \left (\omega t - \dfrac {2\pi x}{\lambda} +\phi \right )$ is

0%
$\dfrac {\lambda \phi}{2\pi}$
0%
$\dfrac{\lambda}{2\pi} (\dfrac {\pi}{2} + \phi)$
0%
$\dfrac{\lambda}{2\pi} (\dfrac {\pi}{2} - \phi)$
0%
$\dfrac {2\pi \phi}{\lambda}$

Explanation

Given,

$y_{1} = a_{1}\sin \left (\omega t - \dfrac {2\pi x}{\lambda}\right )$ and

$y_{2} = a_{2} \cos \left (\omega t - \dfrac {2\pi x}{\lambda} + \phi \right )$

$= a_{2}\sin \left (\dfrac {\pi}{2} + \omega t - \dfrac {2\pi x}{\lambda} + \phi \right )$

$\therefore$ Phase of first wave,

$\phi_{1} = \left (\omega t - \dfrac {2\pi x}{\lambda}\right )$
and phase of second wave,

$\phi_{2} = \dfrac {\pi}{2} + \omega t - \dfrac {2\pi x}{\lambda} + \phi$

$\therefore$ Phase difference,

$\triangle \phi = \phi_{2} - \phi_{1} = \dfrac {\pi}{2} + \phi$

$\therefore$ Path difference

$= \dfrac {\lambda}{2\pi}\times phase\ difference$

$= \dfrac {\lambda}{2\pi}\left (\dfrac {\pi}{2} + \phi \right )$ .

The minimum phase difference between two simple harmonic oscillations,

$y_1 = \dfrac{1}{2} sim \omega t + \dfrac{\sqrt 3}{2} cos \omega t$

$y_2 = sim \omega t + cos \omega t$ , is

0%
$\dfrac{7\pi}{12}$
0%
$\dfrac{\pi}{12}$
0%
$\dfrac{-\pi}{6}$
0%
$\dfrac{\pi}{6}$

Explanation

$y_1 = \dfrac{1}{2} sim \omega t + \dfrac{\sqrt 3}{2} cos \omega t$
Thus we get

$y_1= cos \dfrac{\pi}{3} sim \omega t + sin \dfrac{\pi}{3} cos \omega t$

$\therefore y_1 = sim (\omega t + \pi/3)$

$y_2 = \sqrt 2 \left( \dfrac{1}{\sqrt 2} sim \omega t + \dfrac{1}{\sqrt 2} cos \omega t\right )$
Similarly,

$y_2 = \sqrt 2 sin (\omega t + \pi/4)$
Phase difference

$= \Delta \Phi = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$

A particle performing S.H.M. starts from equilibrium position and its time period is

$16$ second. After

$2$ seconds its velocity is

$\pi \ m/s$ . Amplitude of oscillation is

$(\cos 45^{\circ} = \dfrac {1}{\sqrt {2}})$

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$2\sqrt {2} m$
0%
$4\sqrt {2}m$
0%
$6\sqrt {2} m$
0%
$8\sqrt {2}m$

Explanation

let

$x=asin\omega t$

on differentiating both side

$v=A\omega cos\omega t$

$\pi=A\times\dfrac{2\pi}{T}cos\dfrac{2\pi}{T}t$

on putting T=16 sec and t=2 sec

$8=Acos45$

$A=8\sqrt{2}$

Two sound waves having a phase difference of

$60^o$ , have a path difference of :

0%
$\dfrac{\lambda}{3}$
0%
$\dfrac{\lambda}{6}$
0%
$\dfrac{\lambda}{9}$
0%
$\lambda$

Explanation

Path difference

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

$=\dfrac{\lambda}{2\pi}\times \dfrac{\pi}{3} = \dfrac{\lambda}{6}$

A simple harmonic wave of amplitude

$8$ unit travels along positive

$x$ -axis. At any given instant of time, for a particle at a distance of

$10 cm$ from the origin, the displacement is

$+6$ unit and for a particle at a distance of

$25 cm$ from the origin, the displacement is

$+4$ unit. Calculate the wavelength.

0%
$200 cm$
0%
$230 cm$
0%
$210 cm$
0%
$250 cm$

Explanation

$Y=A\sin { \dfrac { 2\pi }{ \lambda } } \left( vt-x \right)$

$\dfrac { Y }{ A } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { x }{ \lambda } \right)$
In first case,

$\dfrac { { Y }_{ 1 } }{ A } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { { x }_{ 1 } }{ \lambda } \right)$
Here,

${ Y }_{ 1 }=+6$ ,

$A=8$ ,

${ x }_{ 1 }=10 cm$

$\dfrac { 6 }{ 8 } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } \right)$ .....(i)
Similarly in the second case,

$\dfrac { 4 }{ 8 } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } \right)$ .....(ii)
From equation (i),

$2\lambda \left( \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } \right) =\sin ^{ -1 }{ \left( \dfrac { 6 }{ 8 } \right) } =0.85 rad$

$\Rightarrow \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } =0.14$ ....(iii)
Similarly from equation (ii),

$2\pi \left( \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } \right) =\sin ^{ -1 }{ \left( \dfrac { 4 }{ 8 } \right) } =\dfrac { \pi }{ 6 } rad$

$\Rightarrow \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } =0.08$ ....(iv)
Subtracting equation (iv) from equation (iii), we get

$\dfrac { 15 }{ \lambda } =0.06$

$\Rightarrow \lambda =250 cm$

Two particles A and B execute simple harmonic motions of periods T and 5T /They start from mean position. The phase difference between them when the particle A complete one oscillation will be:

0%
$0$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{2\pi}{5}$
0%
$\dfrac{\pi}{6}$

Explanation

Equations of motion of the practicals are

$X_1=A_1 sin \dfrac{2\pi}{T_1}t$

and

$X_2=A_2 sin \dfrac{25}{T_2}t$

$\therefore$ Phase difference

$\Delta \phi =\left ( \dfrac{2\pi}{T_1}-\dfrac{2\pi}{T_2} \right )t$

$=\left ( \dfrac{2\pi}{T}-\dfrac{2\pi}{5T/4} \right )t$

at t=T

$\Delta \phi =\left ( 2 \pi -\dfrac{4 \times 2\pi}{5} \right )\frac{T}{T}=\dfrac{2\pi}{5}$

The phenomena during arising due to the superstition of waves is/are :

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Beats
0%
Stationary waves
0%
Lissajous figures
0%
All of these

Two particles P and Q describe SHM of same amplitude a frequency v along the same straight line. The maximum distance between two particles is

$\displaystyle \sqrt { 2a }$ . The initial phase difference between the particles is :

0%
Zero
0%
$\displaystyle { \pi }/{ 2 }$
0%
$\displaystyle { \pi }/{ 6 }$
0%
$\displaystyle { \pi }/{ 3 }$

Explanation

Let equation of motion to particles be given by,

$x_1=asin(\omega t)$ and

$x_2=asin(\omega t+\phi)$

So initial phase difference between two particles is

$\phi$ .

Now,

$x_2-x_1=a[sin(\omega t+\phi)-sin(\omega t)]=2asin\bigg(\dfrac{\omega t+\phi-\omega t}{2}\bigg)cos\bigg(\dfrac{\omega t+\phi+\omega t}{2}\bigg)$

$=2asin(\phi/2)cos(\omega t+\phi/2)$

Maximum value of above expression is

$2asin(\phi/2)=\sqrt{2}a$

$\implies sin(\phi/2) = 1/\sqrt{2}\implies \phi/2=\pi/4$

$\implies \phi=\pi/2$

A progressive wave is represented as

$y=0.2\cos \pi (0.04 t+0.2x-\dfrac{\pi}{6})$ where distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of

$\dfrac{\pi}{2}$ ?

0%
4 cm
0%
8 cm
0%
25 cm
0%
12.5 cm

Explanation

On comparing the given equation with

$y=a\cos (\omega t + kx - \phi)$
we get,

$k=\dfrac{2\pi}{\lambda}=0.02$
and

$\lambda = 100 cm$
It is given that phase difference between particles

$\triangle \phi = \dfrac{\pi}{2}$
So, the path difference between them,

$\triangle p = \dfrac{\lambda}{2\pi}\times \triangle \phi$

$=\dfrac{\lambda}{2\pi}\times \dfrac{\pi}{2}=\dfrac{\lambda}{4}=\dfrac{100}{4}=25 cm$

With the propagation of a longitudinal wave through a material medium the quantities transmitted in the propagation direction are

0%
Energy, momentum and mass
0%
Energy
0%
Energy and mass
0%
Energy and linear momentum

Explanation

$\textbf{Explanation}$

$\bullet$ In longitudinal wave,particle vibrates in the direction of motion of wave.

$\bullet$ When wave travels it means energy transfers from one end to other in the direction of propagation of wave.

$\bullet$ Particles vibrate in the direction of propagation of wave so linear momentum also transmitted in the direction of propagation of wave.

At a moment in a progressive wave, the phase of a particle executing SHM is

$\dfrac {\pi}{3}$ . Then the place of the particle

$15\ cm$ ahead and at time

$\dfrac {T}{2}$ will be, if the wavelength is

$60\ cm$ .

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

Explanation

Let the phase of second particle is

$\phi$ . Hence the phase difference two particles is

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

$\left (\phi - \dfrac {\pi}{3}\right ) = \dfrac {2\pi}{60}\times 15$

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

$\phi = \dfrac {5\pi}{6}$ .

Equations of a stationary and a travelling waves are as follows,

${ y }_{ 1 }=a\sin { kx } \cos { \omega t }$ and

${ y }_{ 2 }=a\sin { \left( \omega t-kx \right) }$ . The phase difference between two points

${ x }_{ 1 }=\dfrac { \pi }{ 3k }$ and

${ x }_{ 2 }=\dfrac { 3\pi }{ 2k }$ are

${ \phi }_{ 1 }$ and

${ \phi }_{ 2 }$ respectively for two waves. The ratio

$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } }$ is :

0%
$1$
0%
${ 5 }/{ 6 }$
0%
${ 3 }/{ 4 }$
0%
${ 6 }/{ 7 }$

Explanation

${ x }_{ 1 }=\dfrac { \pi }{ 3k }$ and

${ x }_{ 2 }=\dfrac { 3\pi }{ 2k }$

$\sin{k{ x }_{ 1 }}$ or

$\sin{k{ x }_{ 2 }}$ is not zero.

Therefore, neither of

${ x }_{ 1 }$ nor

${ x }_{ 2 }$ is a node.

$\Delta x={ x }_{ 2 }-{ x }_{ 1 }=\left( \dfrac { 3 }{ 2 } -\dfrac { 1 }{ 3 } \right) \dfrac { \pi }{ k } =\dfrac { 7\pi }{ 6k }$

Since,

$\dfrac { 2\pi }{ k } > \Delta x > \dfrac { \pi }{ k }$

$\lambda > \Delta x> \dfrac { \lambda }{ 2 }$

$\left\{ k=\dfrac { 2\pi }{ \lambda } \right\}$

Therefore,

${ \phi }_{ 1 }=\pi$

and

${ \phi }_{ 2 }=k$

$\Delta x=\dfrac { 7\pi }{ 6 }$

Therefore,

$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } } =\dfrac { 6 }{ 7 }$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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