Waves Test

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Waves Test - 22

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Weekly Quiz Competition

Question 1
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A plane progressive wave has frequency $$25 \ Hz$$ and amplitude $$2.5 \times 10^{-5}$$ m and initial phase zero, propagates along the negative x-direction with a velocity of $$300 m/s$$. The phase difference between the oscillations at two points $$6m$$ apart along the line of propagation is :

A
$$\pi $$

B
$$\displaystyle \dfrac{\pi}{2}$$

C
$$2\pi $$

D
$$\displaystyle \dfrac{\pi}{4}$$

Solution

$$t=\dfrac{d}{V}$$ $$=\dfrac{6}{300}$$ $$\Rightarrow t=2\times 10^{-2}$$
$$T=\dfrac{1}{25}$$ $$=4\times 10^{-2}$$
For $$2\pi$$ time taken to $$T$$ and phase difference in $$t$$
$$=\dfrac{2\pi}{T}\times t$$ $$=\dfrac{2\pi}{4\times 10^{-2}}\times 2\times 10^{-2}$$ $$=\pi$$
Question 2
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The equations of displacement of two waves are given as $$Y_{1}=10 \sin(\pi t+\pi /3)$$ $$Y_{2}=5(\sin3\pi t +\sqrt{3}\cos3\pi t$$), then the ratio of their amplitude is

A
$$1:2$$

B
$$2:1$$

C
$$1:1$$

D
$$3:1$$

Solution

Amplitude of $$ { Y }_{ 1 }=10\sin { \left( m\pi t+\pi /3 \right) } $$ is $$ { A }_{ 1 }=10$$
Ampltude of $$ { Y }_{ 2 }=10\left( \dfrac { 1 }{ 2 } \sin { 3\pi t } +\dfrac { \sqrt { 3 } }{ 2 } \cos { 3\pi t } \right) =10\sin { \left( 3\pi t+\pi /3 \right) } $$ is $$ { A }_{ 2}=10$$
$$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =1:1$$
Question 3
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Mark correct option:

A
the phase of transmitted wave always remains unchanged

B
the amplitude of transmitted wave does not depend upon the velocity of wave in media

C
the amplitude of reflected wave and transmitted wave are same to each other for a given incident wave

D
the amplitude of rerflected wave is equal to the amplitude of incident wave

Solution

During the transmission of a wave from one to another media the amplitude of wave changes since it depends on velocities of wave in different media.
But whatever the value of velocities the amplitude of transmission always remain positive so phase remain unchanged.
Question 4
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Three waves of equal frequency having amplitudes $$10 \ mm, 4 \ mm$$ and $$7\ mm$$ arrive at a given point with successive phase difference $$\dfrac \pi 2$$. The amplitude of the resulting wave (in mm) is given by:

A
$$7$$

B
$$6$$

C
$$5$$

D
$$4$$

Solution

It is given that the $$3$$ waves arrive at a point with a successive difference of $$90^o$$.
Therefore, the waves with amplitude $$10 mm$$ and $$7 mm$$ will be completely out of phase and interfere destructively.
Hence, the amplitude of the resultant wave is $$10 - 7 = 3mm$$
Now, the waves of $$3 \ mm$$ and $$4 \ mm$$ are out of phase by $$90^o$$
Therefore, applying Pythagorus theorem, they will interfere with each other causing the new amplitude of the wave to be,
$$ A^2 = 3^2 + 4^2 $$
$$ A^2 = 25 $$ or $$A = 5 \ mm$$
Question 5
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When a transverse plane wave traverses a medium, individual particles execute periodic motion given by the equation $$y=0.25\cos(2\pi t-\pi x)$$. The phase difference for two positions of same particle which are occupied by time intervals $$0.4 second$$ apart is

A
$$144^{o}$$

B
$$135^{o}$$

C
$$72^{o}$$

D
$$108^{o}$$

Solution

$$\omega =\dfrac{2\pi}{T}$$
$$T=\dfrac{2\pi}{\omega}$$
$$T=1sec$$
$$\therefore 2\pi \ rad \ in \ 1 sec$$
$$in \ 0.4 sec \ 2\pi\times 0.9$$
$$=144^o$$
Question 6
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The equation of a wave pulse is given as $$y=\displaystyle \frac{0.8}{(4x+5t)+4}$$, the amplitude of the pulse is:

A
$$0.2 \ units$$

B
$$0.4\ units$$

C
$$0.6 \ units$$

D
$$0.8 \ units$$

Solution

The amplitude of wave is given by $$y$$, at $$x=0$$ and $$t=0$$.
So amplitude is $$y = 0.8/(4\times0 + 5\times0 + 4) = 0.2\ units$$
Question 7
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The displacement of a particle varies according to the relation $$X=4(\cos\pi t+\sin\pi t)$$
The amplitude of the particle is

A
$$-4$$

B
$$4$$

C
$$4\sqrt{2}$$

D
$$8$$

Solution

$$x=4(\cos\ \pi t+\sin\ \pi t)$$
$$ = 4 \sqrt2(\cos \ \pi t \sin \pi /4 + \sin \ \pi t \cos \ \pi /4)$$
$$=4\sqrt{2}\sin (\pi t+\ \dfrac{\pi }{4})$$
So, amplitude $$=4\sqrt{2}$$
Question 8
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The equation of a wave travelling on a stretched string is $$y=4\displaystyle \sin 2\pi(\frac{t}{0.02}-\frac{x}{100})$$ here x, y are in cm and t in seconds. The relative deformation amplitude of medium is

A
0.02$$\pi $$

B
0.08$$\pi $$

C
0.06$$\pi $$

D
0.01$$\pi $$

Solution

$$y=4sin\ 2\pi \left( \dfrac{t}{0.02}-\dfrac

{x}{100} \right)$$
$$\dfrac{dy}{dx}=\dfrac{8\pi}{100}\ \ \ cos\ 2\pi\left( \dfrac{t}{0.02}-\dfrac{x}{100} \right)$$
Deformation amplitude $$A=\left | \dfrac{dy}{dx} \right |=\dfrac{8\pi}{100}$$$$=0.08\pi$$
Question 9
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A wave of length $$2m$$ is superposed on its reflected wave to form a stationary wave. A node is located at $$ x=3m$$ The next node will be located at $$x=$$

A
$$4m$$

B
$$3.75m$$

C
$$3.50m$$

D
$$3.25m$$

Solution

Since wave length is $$2m$$, half of wavelength is $$1m$$. Node forms after each half of wave length. So, node will be formed at each $$1 m$$
So, as node is formed at $$3 m$$
next node will be formed at $$3+1=4\ m.$$
Question 10
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A Sound wave with an amplitude of $$3 cm$$ starts moving towards right from origin and gets reflected at a rigid wall after a second. If the velocity of the wave is $$340 ms^{-1}$$ and it has a wavelength of $$2 m$$, the equations of incident and reflected waves are

A
$$y = 3\times 10^{-2}\sin \pi (340 t - x), y = -3\times 10^{-2}\sin\pi(340t + x)$$ towards left

B
$$y = 4\times 10^{-2} \sin \pi (340 t + x), y = -4\times 10^{-2}\sin\pi(340t + x)$$ towards left

C
$$y = 5\times 10^{-2} \sin \pi (340 t - x), y = -5\times 10^{-2}\sin\pi(340t - x)$$ towards left

D
$$y = 6\times10^{-2}\sin \pi(340 t - x), y = -6\times 10^{-2}\sin\pi(340t + x)$$ towards left

Solution

$$A=3cm$$
$$y=A \sin (Vt-x)$$
$$V= \lambda / t$$
$$f=170 sec$$
$$y_{1}=A \sin (V t-x)$$
$$y_{2}=A \sin (V t+x)$$
$$\therefore y_{1}=3\times 10^{-2} \sin k (340 t-x),$$
$$y_{2}=3\times 10^{-2} \sin k(340 t +x)$$
$$k=\dfrac{2\pi}{\lambda}$$
$$k=\dfrac{2\pi}{2}$$
$$k=\pi$$
$$\therefore y_{1}=3\times 10^{-2} \sin \pi(340t-x)$$
$$y_{2}=3\times 10^{-2} \sin \pi (340t + x)$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 22

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Waves Test - 22

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SHARING IS CARING

Question 1

$$\pi $$

$$\displaystyle \dfrac{\pi}{2}$$

$$2\pi $$

$$\displaystyle \dfrac{\pi}{4}$$

Solution

Question 2

$$1:2$$

$$2:1$$

$$1:1$$

$$3:1$$

Solution

Question 3

the phase of transmitted wave always remains unchanged

the amplitude of transmitted wave does not depend upon the velocity of wave in media

the amplitude of reflected wave and transmitted wave are same to each other for a given incident wave

the amplitude of rerflected wave is equal to the amplitude of incident wave

Solution

Question 4

$$7$$

$$6$$

$$5$$

$$4$$

Solution

Question 5

$$144^{o}$$

$$135^{o}$$

$$72^{o}$$

$$108^{o}$$

Solution

Question 6

$$0.2 \ units$$

$$0.4\ units$$

$$0.6 \ units$$

$$0.8 \ units$$

Solution

Question 7

$$-4$$

$$4$$

$$4\sqrt{2}$$

$$8$$

Solution

Question 8

0.02$$\pi $$

0.08$$\pi $$

0.06$$\pi $$

0.01$$\pi $$

Solution

Question 9

$$4m$$

$$3.75m$$

$$3.50m$$

$$3.25m$$

Solution

Question 10

$$y = 3\times 10^{-2}\sin \pi (340 t - x), y = -3\times 10^{-2}\sin\pi(340t + x)$$ towards left

$$y = 4\times 10^{-2} \sin \pi (340 t + x), y = -4\times 10^{-2}\sin\pi(340t + x)$$ towards left

$$y = 5\times 10^{-2} \sin \pi (340 t - x), y = -5\times 10^{-2}\sin\pi(340t - x)$$ towards left

$$y = 6\times10^{-2}\sin \pi(340 t - x), y = -6\times 10^{-2}\sin\pi(340t + x)$$ towards left

Solution

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