Waves Test

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Waves Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Phase of a particle gives us

A
the velocity of the particle

B
the direction of motion of the particle

C
the acceleration of the particle

D
the displacement of the particle

Solution

phase is the relative position of the particle w.r.t the origin , Thus, it can also be used to find the direction of motion of the particle at a given time

The correct option is (b)
Question 2
1 / -0

Which of the following functions represent a wave

A
$$(x-vt)^{2}$$

B
$$\log (x+vt)$$

C
$$e^{-(x-vt)^{2}}$$

D
$$\dfrac{1}{x+vt}$$

Solution

Although all the four function are written in the form $$f\left(ax \pm bt\right),$$ only option $$(c)$$ among the four function is finite everywhere at all times. Hence only option $$(c)$$ represents a wave.
Question 3
1 / -0

The equation of a wave traveling in a string can be written as $$y=3 \sin\pi(100t-x)$$cm. Its wavelength is:

A
3 cm

B
100 cm

C
5 cm

D
2 cm
Question 4
1 / -0

Three waves $$y _ { 1 } = 2 A \sin \left( \omega t - \dfrac { \pi } { 3 } \right)$$ $$y _ { 2 } = 2 A \sin \left( \omega t + \dfrac { \pi } { 3 } \right)$$ and $$y _ { 3 } = - A \sin ( \omega t )$$ interfere each other. The amplitude of the resultant wave is

A
A

B
zero

C
2 A

D
3 A

Solution

Amplitude of resultant wave
$$y=y_1+y_2+y_3$$
$$y=2A\sin \left(\omega t-\dfrac{\pi}{3}\right)+2A\sin\left(\omega t+\dfrac{\pi}{3}\right)-A\sin\omega t$$
$$y=2A\sin\omega t\cos\dfrac{\pi}{3}-2A\cos \omega t\sin\dfrac{\pi}{3}+2A\sin\omega t\cos \dfrac{\pi}{3}+2A\cos \omega t\sin\dfrac{\pi}{3}-A\sin\omega t$$
$$y=4A\sin\omega t\cos \dfrac{\pi}{3}-A\sin\omega t$$
$$y=2A\sin\omega t-A\sin\omega t$$
$$y=A\sin\omega t$$
Thus, the resultant amplitude is A.
Question 5
1 / -0

If two waves of same frequency and amplitude superpose to produce resultant of the amplitude of either wave, then their phase difference is-

A
$$\pi$$

B
$$2\pi/3$$

C
$$\pi/3$$

D
$$Zero$$

Solution

Let,
$$\begin{array}{l} { x_{ 1 } }=A\sin wt \\ { x_{ 2 } }=A\sin \left( { wt+\phi } \right) \\ x={ x_{ 1 } }+{ x_{ 2 } } \end{array}$$
$$\therefore$$ For resultant amplitude to be equal to
A $$\phi $$ must be $$\dfrac{{2\pi }}{3}$$.
$$\therefore$$ phase difference is $$\dfrac{{2\pi }}{3}$$
$$\therefore$$ Option $$B$$ is correct answer.
Question 6
1 / -0

The displacement of a particle executing SMH is given by $$x=0.01{\,\,}sin{\,\,}100\pi(t+0.05).$$ The time period is

A
$$0.01$$sec

B
$$0.02$$sec

C
$$0.1$$sec

D
$$0.2$$sec

Solution

Given, Angular frequency $$(\omega) = 100\pi$$

So, Time period $$ = \dfrac{2\pi}{\omega} = \dfrac{2 \pi}{100 \pi} = 0.02 \;sec$$

Therefore, B is correct option.
Question 7
1 / -0

The phase difference between the displacement and acceleration of particle executing S.H.M. radian is:

A
$$ \pi / 4 $$

B
$$ \pi / 2 $$

C
$$ \pi $$

D
$$ 2 \pi $$

Solution

$$va\sin(\omega t+\phi)$$
Differentiating
$$v=a\omega\cos(\omega t+\phi)$$
Differentiating again,
$$a=-a\omega^2\sin(\omega+\phi)$$
Now taking minus sign
$$a=a\omega^2\sin(\omega t+\phi+\pi)$$
So, phase difference between displacement and acceleration will be:-
$$=\omega t+\phi+\pi-{\omega t+\phi}\\=\omega t+\phi+\pi-\omega t -\phi\\=\pi$$
Question 8
1 / -0

The two waves represented by $$y=a \sin (\omega t)$$ and $$y_2=b \cos (\omega t)$$ have a phase difference of

A
$$0$$

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

C
$$\pi$$

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

Solution

$$y_1=a \sin \omega t$$, and $$y_2=b \cos \omega t=b \sin \left(\omega t+\dfrac{\pi}{2}\right)$$
So phase difference $$\phi=\pi/2$$
Question 9
1 / -0

The displacement of a particle is given by$$x = 3 sin(5\pi t) + 4 cos 5(5 \pi t)$$ The amplitude of the particle is [MP PMT 1999]

A
3

B
4

C
5

D
7

Solution

For the given imposing imposing waves
$$a_1 = 5, a_2 = 4 $$ and phase difference $$\phi = \dfrac{\pi}{2}$$
$$\Rightarrow A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 cos \pi /2} = \sqrt{(3)^2 + (4)^2} = 5$$
Question 10
1 / -0

The frequency of light ray having the wavelength $$3000\ A^o$$ is

A
$$9 \times 10^{13}\ cycles/sec$$

B
$$ 10^{15}\ cycles/sec$$

C
$$90 \ cycles/sec$$

D
$$3000 \ cycles/sec$$

Solution

The frequency of the light ray is given as:
$$\nu=\dfrac{c}{\lambda}$$

$$=\dfrac{3 \times 10^8}{3000 \times 10^{-10}}$$

$$=10^{15}\ cycles/sec$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 20

Result

Waves Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

the velocity of the particle

the direction of motion of the particle

the acceleration of the particle

the displacement of the particle

Solution

Question 2

$$(x-vt)^{2}$$

$$\log (x+vt)$$

$$e^{-(x-vt)^{2}}$$

$$\dfrac{1}{x+vt}$$

Solution

Question 3

3 cm

100 cm

5 cm

2 cm

Question 4

A

zero

2 A

3 A

Solution

Question 5

$$\pi$$

$$2\pi/3$$

$$\pi/3$$

$$Zero$$

Solution

Question 6

$$0.01$$sec

$$0.02$$sec

$$0.1$$sec

$$0.2$$sec

Solution

Question 7

$$ \pi / 4 $$

$$ \pi / 2 $$

$$ \pi $$

$$ 2 \pi $$

Solution

Question 8

$$0$$

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

$$\pi$$

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

Solution

Question 9

3

4

5

7

Solution

Question 10

$$9 \times 10^{13}\ cycles/sec$$

$$ 10^{15}\ cycles/sec$$

$$90 \ cycles/sec$$

$$3000 \ cycles/sec$$

Solution

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