Waves Test

Result

Waves Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The displacement of a particle in SHM is $$x =3 sin(20\pi t) +4 cos (20\pi t)$$ cm.Its amplitude of oscillation is

A
3 cm

B
4 cm

C
5 cm

D
25 cm

Solution

Both the SHM have same frequency ,with an amplitude of $$3$$ and $$4$$, with a phase difference of $$90^0$$

The maximum value of the equation,
$$x=3\sin(20\pi t)+4\cos(20\pi t)$$ is

$$\sqrt{3^2+4^2}= 5\ cm$$

Option $$\textbf C$$ is the correct answer
Question 2
1 / -0

A progressive wave moves with a velocity of $$36m/s$$ in a medium with a frequency of $$200Hz$$. The phase difference between two particles separated by a distance of $$1cm$$ is:

A
$${40 }$$

B
$$20$$

C
$$\dfrac{\pi }{9}$$

D
$$\dfrac{{{\pi }}}{16}$$

Solution

Solution we have, $$ V = 36m/s, f = 200H_{2} $$ and
$$ \Delta x = 1cm = 10^{-2}m $$
we know that,
$$ v = f\lambda $$
$$ \Rightarrow \lambda = \dfrac{v}{f} = \dfrac{36}{200} = \dfrac{9}{50}m $$
Also, $$ \Delta \phi = 2\pi (\dfrac{\Delta x}{\lambda }) $$
$$ \Rightarrow \Delta \phi = 2\pi \dfrac{(\dfrac{1}{100})}{\dfrac{9}{50}} \Rightarrow \Delta \phi = \dfrac{\pi }{4} $$
Option - c is correct
Question 3
1 / -0

Two identical sinusoidal waves each of amplitude $$5 mm$$ with a phase difference of $$\dfrac{\pi}{2}$$ are travelling in the same direction in a string. The amplitude of the resultant wave (in mm) is:-

A
Zero

B
$$5 \sqrt{2}$$

C
$$\dfrac{5} {\sqrt{2}}$$

D
$$2.5$$

Solution

Given amplitude = 5mm
So, sinusoidal wave can be
taken as $$ A = 5 sin\omega t $$
Another wave has $$ \dfrac{\pi}{2}$$ phase diff-
$$ \Rightarrow B = 5 sin(\omega t+\dfrac{\pi}{2})$$
$$ = 5 cos \omega t $$
$$ A+B = 5(sin\omega t + cos \omega t )$$
$$= 5\sqrt{2}(\dfrac{1}{\sqrt{2}}sin\omega t + \dfrac{1}{\sqrt{2}} cos\omega t)$$
$$ = 5\sqrt{2} (sin\omega t \,cos 45^{\circ}+cos \omega t\, sin45^{\circ})$$
$$ = 5\sqrt{2} (sin(\omega t +45^{\circ}))$$
$$ = 5\sqrt{2} sin (\omega t+ \pi/4)$$
Question 4
1 / -0

$$x_1 = A\sin(\omega t - 0.1x)$$ and $$x_2 = A\sin(\omega t - 0.1x - \dfrac{\phi}{2})$$. The resultant amplitude of combined wave is:-

A
$$2 A\cos \dfrac{\phi}{4}$$

B
$$A \sqrt{2\cos \dfrac{\phi}{2}}$$

C
$$2 A\cos \dfrac{\phi}{2}$$

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

Solution

$$ x_1 = A sin (\omega t- 0.1 x)$$
$$ x_2 = a sin(\omega t-01x-\phi /2)$$
$$ x_1+x_2 = A(sin(\omega t-0.1x)+sin(\omega t-0.1x-\phi /2))$$
$$ = A (2sin \dfrac{(\omega t-0.1x)+(\omega t-0.1x-\phi /2)}{2}$$
$$ cos \dfrac{(\omega t-0.1x)-(wt-01x-\phi /2)}{2})$$
$$ = (2A cos \phi /4) (sin(\omega t-0.1x - \phi /2))$$
Amplitude = $$ 2A cos \phi /4$$
Question 5
1 / -0

The value of phase, at maximum displacement from the mean position of a particle, in SHM is ?

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

B
$$\pi $$

C
zero

D
$$2\pi $$

Solution
Question 6
1 / -0

The phase difference between two waves represented by
$${y_1} = {10^{ - 6}}\sin \left[ {100t + \frac{x}{{50}} + 0.5} \right]m$$
$${y_2} = {10^{ - 6}}\cos \left[ {100t + \frac{x}{{50}}} \right]m$$
where x is expressed in metres and t is expressed in seconds is approximately

A
1.07 rad

B
2.07 rad

C
0.5 rad

D
1.5 rad

Solution

$$y_{1}=10^{-6} \sin \left[100 t + \dfrac{x}{50}+ 0.5\right] m$$
$$y_{2}=10^{-6} \cos \left[100 t + \dfrac{x}{50}\right] m$$
So,$$y^{1}=10^{-6} \cos\left[\dfrac{\pi}{2}-\left(100 t + \dfrac{x}{50}+0.5\right)\right]$$
$$y^{1}=10^{-6} \cos \left[\dfrac{-\pi}{2}+100 t+\dfrac{x}{50}+0.5\right]$$
$$\triangle \phi =\phi_{2}-\phi_{1}$$
$$\phi_{2}=0$$ where as $$\phi_{1}=0.5-\dfrac{\pi}{2}$$
So,
$$\triangle \phi=\phi_{2}-\phi_{1}=\dfrac{\pi}{2}-0.5=1.07 \ rad$$
option $$-A$$ is correct.
Question 7
1 / -0

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.

A
$$\dfrac{\lambda }{{2\theta }}$$

B
$$\dfrac{2\lambda }{{\theta }}$$

C
$$\dfrac{\lambda }{{\theta }}$$

D
$$2 \lambda \ \sin \theta $$

Solution
Question 8
1 / -0

The amplitude of a particle due to superposition of following S.H.Ms. Along the same line is
$${ X }_{ 1 }=2sin50\pi t\quad ;\quad { X }_{ 2 }=10sin\left( 50\pi t+{ 37 }^{ } \right) $$
$${ X }_{ 3 }=-4sin50\pi t\quad ;\quad { X }_{ 4 }=-12sin50\pi t$$

A
$$4\sqrt { 2 } $$

B
$$4$$

C
$$6\sqrt { 2 } $$

D
None of these

Solution
Question 9
1 / -0

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

A
$$ 2 : 1, 0 $$

B
$$ 1 : 2, 0 $$

C
$$ 1 : 2, \pi /2 $$

D
$$ 2 : 1, \pi / 2 $$

Solution

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.
Question 10
1 / -0

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

A
$$40 ^ { \circ }$$

B
$$20 \mathrm { rad }$$

C
$$\dfrac { \pi } { 9 } \mathrm { rad }$$

D
$$\dfrac { \pi } { 9 } ^{o}$$

Solution