Waves Test

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Waves Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Two particles are executing $$S.H.M$$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?

A
$$5\pi/6$$

B
$$2\pi/3$$

C
$$\pi/3$$

D
$$\pi/6$$

Solution

$$\quad y = a\sin(\omega t + \phi); \mbox{ when } y = a/2 \\ \quad then\space \displaystyle\frac{a}{2} = a\sin(\omega t + \phi)$$

$$\quad or \space \sin(\omega t + \phi) = \displaystyle\frac{1}{2} = \sin\displaystyle\frac{\pi}{6} \space or \space \sin\displaystyle\frac{5\pi}{6}$$

So phase of two particles is $$\pi/6$$ and $$5\pi/6$$ radians.

Hence phase difference $$ = (5\pi/6) - \pi/6 = 2\pi/3$$
Question 2
1 / -0

The equation $$ y=A sin^2(kx-\omega t) $$ represents a wave with

A
amplitude A frequency $$\omega/2\pi $$

B
amplitude A /2 frequency $$\omega/\pi $$

C
amplitude 2A frequency $$\omega/4\pi $$

D
It does not represent a wave motion

Solution

$$y=A\sin^2(kx-\omega t) = \dfrac{A}{2}\{1-\cos(2kx-2\omega t)\}$$
Comparing the above equation with standard progressive wave equation
$$y=A\sin(kx-\omega t)$$, we get

Frequency $$\nu'=\dfrac{\omega'}{2\pi}= \dfrac{2\omega}{2\pi}=\dfrac{\omega}{\pi}$$ and amplitude $$A'=A/2$$
Ans: B
Question 3
1 / -0

The phenomenon of interference is shown by

A
longitudinal mechanical waves only

B
transverse mechanical waves only

C
non-mechanical transverse waves only

D
All of the above

Solution
Question 4
1 / -0

The equation of a plane progressive wave is $$ y=0.9 sin 4\pi[t-\dfrac{x}{2}] $$. When it is reflected at a rigid support, its amplitude becomes $$\dfrac{2}{3} $$ of its previous value. The equation of the reflected wave is

A
$$ y=0.6 sin 4\pi[t+\dfrac{x}{2}] $$

B
$$ y=-0.6 sin 4\pi[t+\dfrac{x}{2}] $$

C
$$ y=-0.9 sin 8\pi[t-\dfrac{x}{2}] $$

D
$$ y=-0.9 sin 4\pi[t+\dfrac{x}{2}] $$

Solution

Amplitude of reflected wave $$=0.9 \times \dfrac{2}{3}=0.6$$
It would travel along negative direction of x-axis , and on reflection at a rigid support, there occurs a phase change of $$\pi$$.
Hence, the wave equation becomes:
$$y = 0.6 \sin 4\pi(t+\dfrac{x}{2}+\pi)$$
$$y =- 0.6 \sin 4\pi(t+\dfrac{x}{2})$$
Question 5
1 / -0

A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second. What is the velocity of ocean waves?

A
2.5 m/s

B
5.0 m/s

C
8.0 m/s

D
10.0 m/s

Solution

Here wavelength of the waves is given as

$$\lambda = 5.0 $$ m.

Frequency $$\nu =2$$

Wave velocity $$v = \nu \lambda= 10$$ m/s.
Question 6
1 / -0

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

A
$$\pi $$

B
$$2\pi/3$$

C
$$zero$$

D
$$\pi/3$$

Solution

We know that:
$$R^2= a^2+b^2+2 ab cos\phi $$
$$\therefore a^2=a^2+a^2+2a^2 cos\phi $$
$$\Rightarrow cos\phi =-\dfrac{1}{2}$$

$$\Rightarrow \phi=\dfrac{2\pi}{3} $$
Question 7
1 / -0

Statement 1: A transverse wave are produced in a very long string fixed at one end. Only progressive wave is observed near the free end.
Statement 2: Energy of reflected wave does not reach the free end

A
Statement-1 is false, Statement -2 is true

B
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for statement-1

D
Statement-1 is true, Statement-2 is false

Solution

Since the string is fixed only at one end, there will be no reflected wave and hence no standing waves will be produced. The reason is that no reflected wave means no reflected energy and hence no interference and no standing waves. Hence both statements are correct and statement 2 is correct explanation of statement 1.
Question 8
1 / -0

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
$$ -4 m $$

B
$$ 4 m $$

C
$$ 4\sqrt{2} m $$

D
$$ 8m $$

Solution

$$x=4(cos \pi t+sin \pi t) $$
$$= 4[sin(\dfrac{\pi}{2}-\pi t)+ sin \pi t ]$$
$$= 8[ sin\dfrac{\pi}{4}. cos(-\dfrac{\pi}{4}+\pi t)] $$ since $$ [cos(-\theta)= cos\theta] $$
$$=\dfrac{8}{\sqrt{2}}. cos[\pi t- \dfrac{\pi}{4}] \\ = 4\sqrt{2} cos[\pi t- \dfrac{\pi}{4}] $$

Comparing it with standard equation:
$$x = A cos(\omega t- kx) \Rightarrow A=4\sqrt{2} $$
Question 9
1 / -0

A particle executing simple harmonic motion along y-axis has its motion described by the equation $$y = A\sin(\omega t) + B$$. The amplitude of the simple harmonic motion is:

A
$$A$$

B
$$B$$

C
$$A+B$$

D
$$\sqrt{A+B}$$

Solution

The equation $$y = A sin (wt) + B$$ suggests that the particle is executing Simple Harmonic Motion about $$B$$ which is the mean position and the amplitude of oscillation of the SHM (about $$B$$) is equal to $$A$$.
Question 10
1 / -0

Sound waves of wavelength $$\lambda $$ travelling with velocity $$v$$ in a medium enter into another medium in which their velocity is $$4v$$. The wavelength in $$2^{nd}$$ medium is :

A
$$4\lambda$$

B
$$\lambda $$

C
$$\lambda/4 $$

D
$$ 16\lambda $$

Solution

From $$v=n\lambda $$ we find $$\lambda\propto v$$ because frequency n is constant .
Therefore ,
new wavelength = $$4\lambda $$

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Re-Attempt Weekly Quiz Competition

Waves Test - 30

Result

Waves Test - 30

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Question 1

$$5\pi/6$$

$$2\pi/3$$

$$\pi/3$$

$$\pi/6$$

Solution

Question 2

amplitude A frequency $$\omega/2\pi $$

amplitude A /2 frequency $$\omega/\pi $$

amplitude 2A frequency $$\omega/4\pi $$

It does not represent a wave motion

Solution

Question 3

longitudinal mechanical waves only

transverse mechanical waves only

non-mechanical transverse waves only

All of the above

Solution

Question 4

$$ y=0.6 sin 4\pi[t+\dfrac{x}{2}] $$

$$ y=-0.6 sin 4\pi[t+\dfrac{x}{2}] $$

$$ y=-0.9 sin 8\pi[t-\dfrac{x}{2}] $$

$$ y=-0.9 sin 4\pi[t+\dfrac{x}{2}] $$

Solution

Question 5

2.5 m/s

5.0 m/s

8.0 m/s

10.0 m/s

Solution

Question 6

$$\pi $$

$$2\pi/3$$

$$zero$$

$$\pi/3$$

Solution

Question 7

Statement-1 is false, Statement -2 is true

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for statement-1

Statement-1 is true, Statement-2 is false

Solution

Question 8

$$ -4 m $$

$$ 4 m $$

$$ 4\sqrt{2} m $$

$$ 8m $$

Solution

Question 9

$$A$$

$$B$$

$$A+B$$

$$\sqrt{A+B}$$

Solution

Question 10

$$4\lambda$$

$$\lambda $$

$$\lambda/4 $$

$$ 16\lambda $$

Solution

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