Waves Test

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Waves Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

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

A
2A

B
A

C
A $\sqrt{2}$

D
A/2

Solution

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)
Question 2
1 / -0

Principle of superposition can be applied to

A
EM waves

B
sound waves

C
Radio waves

D
All of the above

Solution

Superposition of waves applies to all types of waves that are Fourier exandable waves. The waves depicted here are fourier expandable

The correct option is (d)
Question 3
1 / -0

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

A
$2 \pi$

B
$2 \pi/3$

C
$2 \pi/5$

D
$\pi/6$

Solution

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)
Question 4
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
$-4$

B
$4$

C
$4\sqrt2$

D
$8$

Solution

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 $4\sqrt{2}$
Question 5
1 / -0

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

A
$3 \pi/5$

B
$6 \pi/5$

C
$2 \pi/5$

D
$7 \pi/5$

Solution

Phase difference = $(2\pi/5)$ path difference $\implies$ phase difference = $6 \pi/5$

The correct option is (b)
Question 6
1 / -0

If the phase difference between two component waves of different amplitudes is $2\pi$ , their resultant amplitude will become

A
sum of the amplitudes

B
difference of the amplitudes

C
product of the amplitudes

D
ratio of their amplitudes

Solution

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.

If $\theta = 2\pi$ , then $R = A+B$

The correct option is (a)
Question 7
1 / -0

The maximum potential energy / length increases with:

A
Amplitude

B
Wavelength

C
Frequency

D
Velocity

Solution

If $y= A \sin (\omega t- kx)$ is the equation of a wave through a string, then the slope of the wave is $\dfrac{dy}{dx}=- Ak \cos (\omega t- kx)$ .

The maximum potential energy will be $T \times \Delta x \times (\dfrac{dy}{dx})^2A^2k^2=A^2k^2 \cos ^2(\omega t - kx)T \Delta x$

The maximum potential energy will be obtained if cos (\omega t - kx)=1. Thus, maximum potential energy = $4 \pi^2A^2T f \times T \times \Delta x$ ; T is the tension in the string

We also know that $T=\mu v^2$ . Substituting, we get,

Maximum potential energy = $4 \pi^2 f^2A^2 \mu$
Thus maximum potential energy depends on frequency and as frequency increases, potential energy also increases

The correct option is (c)
Question 8
1 / -0

The refracted and the incident pulses for a wave travelling in a string have an amplitude ratio of 1:2. The ratio of their phases will be

A
1:2

B
2:1

C
1:1

D
1:4

Solution

The incident and refracted pulses have same phase. Hence the ratio of phases is 1:1

The correct option is (c)
Question 9
1 / -0

Reflection at a free boundary implies

A
restoring force is zero

B
restoring force is not zero

C
applied force is zero

D
applied force is not zero

Solution

In a free boundary, the restoring force is always zero

The correct option is (a)
Question 10
1 / -0

The initial phase of a pulse travelling on a string is $\pi/3$ . The reflected pulse has a phase of

A
$4 \pi/3$

B
$2 \pi/3$

C
$\pi/3$

D
$5 \pi/3$

Solution

The incident and reflected pulse are $\pi$ radians out of phase. Thus reflected ray will have a phase of $\pi + \pi/3=4 \pi/3$

The correct option is (a)

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Re-Attempt Weekly Quiz Competition

Waves Test - 47

Result

Waves Test - 47

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SHARING IS CARING

Question 1

2A

A

A2\sqrt{2}2​

A/2

Solution

Question 2

EM waves

sound waves

Radio waves

All of the above

Solution

Question 3

2π2 \pi2π

2π/32 \pi/32π/3

2π/52 \pi/52π/5

π/6 \pi/6π/6

Solution

Question 4

−4-4−4

444

424\sqrt242​

888

Solution

Question 5

3π/53 \pi/53π/5

6π/56 \pi/56π/5

2π/52 \pi/52π/5

7π/57 \pi/57π/5

Solution

Question 6

sum of the amplitudes

difference of the amplitudes

product of the amplitudes

ratio of their amplitudes

Solution

Question 7

Amplitude

Wavelength

Frequency

Velocity

Solution

Question 8

1:2

2:1

1:1

1:4

Solution

Question 9

restoring force is zero

restoring force is not zero

applied force is zero

applied force is not zero

Solution

Question 10

4π/34 \pi/34π/3

2π/32 \pi/32π/3

π/3 \pi/3π/3

5π/35 \pi/35π/3

Solution

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A $\sqrt{2}$

$2 \pi$

$2 \pi/3$

$2 \pi/5$

$\pi/6$

$-4$

$4$

$4\sqrt2$

$8$

$3 \pi/5$

$6 \pi/5$

$2 \pi/5$

$7 \pi/5$

$4 \pi/3$

$2 \pi/3$

$\pi/3$

$5 \pi/3$