Waves Test

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Waves Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Which of these does not represent a wave equation

A
$$y=A sin (\omega t- kx)$$

B
$$y=A sin (\omega t + kx)$$

C
$$y=A cos (\omega t- kx)$$

D
$$y=A (\omega t- kx)$$

Solution

For a equation to be a wave equation, it should also be periodic. Adding a period T to any of the equations given above, will give back the same equation, except for the equation depicted in (d)

Thus , option (d) is the right answer
Question 2
1 / -0

A travelling wave is represented by the equation $$y=20 sin 2(x-4t)$$. If y is measured in cms, what is the amplitude of the wave

A
20 mms

B
20 cm

C
20 m

D
2 cm

Solution

From dimensional analysis, we see A takes the same unit as that of y, since trigonometric function doesent have any unit

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

If the particle was to start at the extreme position i.e x = +A, then what will be the equation of SHM:

A
$$A cos (\omega t)$$

B
$$A sin (\omega t)$$

C
$$A cot (\omega t)$$

D
$$A tan (\omega t)$$

Solution

The equation of SHM, when the particle started at the mean position at t = 0 is given by $$x=A sin (\omega t + \phi); \phi $$ is the phase constant.

At t =0, if x=A, then, we have $$sin \phi =1$$ or $$\phi=\pi/2$$

Thus, the equation of SHM will be $$x=A sin(\omega t + \pi/2)=A cos (\omega t)$$

The correct option is (a)
Question 4
1 / -0

A particle executes SHM given by an equation $$x=A sin (\omega t + \pi/6)$$. Where was the particle at t = 0

A
A

B
Mean position

C
A/2

D
A/4

Solution

Putting t = 0, we get x = A/2. Thus, the particle was halfway away from the mean position at t=0

The option (c) is the correct option
Question 5
1 / -0

The equation of a travelling wave is given by $$3sin(\pi/2)(25t-x)$$ where x and y are in meters and t is in seconds. What is the ratio of maximum amplitude to its wavelength

A
1/4

B
3/2

C
3/4

D
1/2

Solution

$$y = 3 sin (\pi/2) (50t - x)$$

The wavelength is $$\lambda=(2 \pi/k)=2 \pi/(\pi/2)=4$$

The maximum amplitude is 3. Hence ratio = 3/4

The correct answer is (c)
Question 6
1 / -0

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude $$A_1$$. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude $$A_2$$. The ratio of $$\left( \frac{A_1}{A_2}\right)$$ is:

A
$$\dfrac{M}{m}$$

B
$$\left(\dfrac{M}{M+m}\right)^{1/2}$$

C
$$ (M+m)\sqrt{\dfrac{k}{M+m}A_2}$$

D
$$\dfrac{M}{M+m}$$

Solution

$$COM \Rightarrow M \sqrt{\dfrac{k}{M}A_1}= (M+m)\sqrt{\dfrac{k}{M+m}A_2}$$
Question 7
1 / -0

A magnifier lens focusses a beam of light on a sheet of paper for 5 secs and the paper starts burning. The process described here is

A
Concentrated light produces heat

B
Light energy is converted in to heat energy

C
Photons heating the paper

D
Electrons heating the paper

Solution

When light is incident on a paper, the ight energy is converted in to heat energy, thus burning the paper

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

Tapping gently on the surface of water produces concentric ripples. This is due to

A
circular interference of waves

B
linear interference of waves

C
energy transfer radially

D
vibration of nearby particles

Solution

Tapping gently on the surface of water produces concentric ripples, because the energy given in the tapping spreads radially along the surface of water and hence the locus of all points having same disturbance is seen as a circle

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

A plane wave of wavelength $$\lambda$$ is incident at an angle $$\theta$$ on a plane mirror. Maximum intensity will be observed at $$P$$, when

A
$$\cos \theta = 3\lambda / 2d$$

B
$$\sec \theta - \cos \theta = 3\lambda / 4d$$

C
$$\cos \theta = \lambda / 4d$$

D
$$\sec \theta - \cos \theta = \lambda / 4d$$

Solution

for max intensity
$$CO+OP=$$multiple of $$\cfrac { \lambda }{ 2 } $$
from fig.
$$\cfrac { dcos{ 2\theta } }{ cos{ \theta } } +\cfrac { d }{ cos{ \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cfrac { d(1+cos{ 2\theta }) }{ \cos { \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cfrac { d2\cos { ^{ 2 }\theta } }{ \cos { \theta } } =\cfrac { \lambda }{ 2 } \\ \Rightarrow \cos { \theta } =\cfrac { \lambda }{ 4d } $$
Question 10
1 / -0

Two particles are executing simple harmonic motion of the same amplitude A and frequency $$\omega$$ along the x-axis. Their mean position is separated by distance $$X_0(X_0>A)$$. If the maximum separation between them is ($$ X_0+ A$$), the phase difference between their motion is:

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{6}$$

D
$$\dfrac{\pi}{2}$$

Solution

$$x1=Asin \big( \omega t + \phi 1\big) $$
$$x2=Asin \big( \omega t + \phi 2\big) $$
$$x1-x2=A \big(2sin \big( \omega t+ \frac{ \phi1 + \phi2 }{2} \big)sin \big( \frac{ \phi1 - \phi2}{2} \big) \big) $$
$$A=2Asin \frac{ \phi1- \phi2 }{2}$$

$$ \frac{ \phi1- \phi2 }{2}= \frac{ \Pi }{6} $$
$$\phi1=\frac{ \Pi }{3}$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 46

Result

Waves Test - 46

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SHARING IS CARING

Question 1

$$y=A sin (\omega t- kx)$$

$$y=A sin (\omega t + kx)$$

$$y=A cos (\omega t- kx)$$

$$y=A (\omega t- kx)$$

Solution

Question 2

20 mms

20 cm

20 m

2 cm

Solution

Question 3

$$A cos (\omega t)$$

$$A sin (\omega t)$$

$$A cot (\omega t)$$

$$A tan (\omega t)$$

Solution

Question 4

A

Mean position

A/2

A/4

Solution

Question 5

1/4

3/2

3/4

1/2

Solution

Question 6

$$\dfrac{M}{m}$$

$$\left(\dfrac{M}{M+m}\right)^{1/2}$$

$$ (M+m)\sqrt{\dfrac{k}{M+m}A_2}$$

$$\dfrac{M}{M+m}$$

Solution

Question 7

Concentrated light produces heat

Light energy is converted in to heat energy

Photons heating the paper

Electrons heating the paper

Solution

Question 8

circular interference of waves

linear interference of waves

energy transfer radially

vibration of nearby particles

Solution

Question 9

$$\cos \theta = 3\lambda / 2d$$

$$\sec \theta - \cos \theta = 3\lambda / 4d$$

$$\cos \theta = \lambda / 4d$$

$$\sec \theta - \cos \theta = \lambda / 4d$$

Solution

Question 10

$$\dfrac{\pi}{3}$$

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

$$\dfrac{\pi}{6}$$

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

Solution

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