Waves Test

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Waves Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is correct?

A
Both, sound and light waves in air are longitudinal.

B
Both, sound and light waves in air are transverse.

C
Sound waves in air are transverse and light waves are longitudinal.

D
Sound waves in air are longitudinal and light waves are transverse.

Solution

sound waves are longitudinal waves and light being a EM wave is transverse in nature .
so the answer is D.
Question 2
1 / -0

A longitudinal wave has a compression to compression distance of 10 m. It takes the wave 5 s to pass a point. Find the velocity of wave.

A
2 m/s

B
3 m/s

C
4 m/s

D
6 m/s

Solution

The speed of wave is given by

$$v = \nu \lambda$$

where, v is the speed of the wave,$$\nu$$ is the frequency of wave and
$$\lambda$$ is its wavelength.

But, $$\nu = \dfrac{1}{T}$$

where T is a period of propagation of the wave.
Hence,

$$v = \dfrac{\lambda}{T}$$

$$v = \dfrac{10}{5} = 2 ms^{-1}$$
Question 3
1 / -0

What happens when a sound wave is reflected from the boundary of a denser medium? The compression of the incident wave is returned as a

A
rarefaction

B
crest

C
trough

D
compression

Solution

Sound wave goes through $$\pi$$ phase change when reflected from the boundary of a denser medium. Hence, the compression of the incident wave is returned as a compression.
Question 4
1 / -0

A particle performing SHM is found at its equilibrium at t=1sec, and it is found to have a speed of 0.25 m/s at t=2 sec. If the period of oscillation is 6 sec , calculate amplitude of oscillation.

A
$$\displaystyle\frac{3}{2\pi}m$$

B
$$\displaystyle\frac{3}{4\pi}m$$

C
$$\displaystyle\frac{6}{\pi}m$$

D
$$\displaystyle\frac{3}{8\pi}$$

Solution

Since a SHM can be represented by $$x=A\sin { \left( \omega t+\phi \right) } $$

$$\Rightarrow x=A\sin { \left( \left( \dfrac { 2\pi }{ T } \right) t+\phi \right) } $$ ; $$T=6s$$

So the equation becomes: $$x=A\sin { \left( \dfrac { \pi t }{ 3 } +\phi \right) } $$

We are given that at $$t=1s, x=0 $$; so we have:
$$0=A\sin

{ \left( \dfrac { \pi }{ 3 } +\phi \right) } \\ \Rightarrow \left(

\dfrac { \pi }{ 3 } +\phi \right) =0\\ \Rightarrow \phi =-\dfrac { \pi

}{ 3 } \\ \Rightarrow x=A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac {

\pi }{ 3 } \right) } \\ \Rightarrow v=\dfrac { dx }{ dt } =\dfrac { d

}{ dt } \left( \sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi }{ 3 }

\right) } \right) =\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi t }{

3 } -\dfrac { \pi }{ 3 } \right) } $$

Now we are given that at $$t=2s, \left| v \right| =0.25m/s$$,
So we have:
$$0.25=\left|

\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { 2\pi }{ 3 } -\dfrac { \pi

}{ 3 } \right) } \right| \\ \Rightarrow 0.25=\left| \dfrac { \pi A }{ 3

} \cos { \left( \dfrac { \pi }{ 3 } \right) } \right| \\ \Rightarrow

0.25=\dfrac { \pi A }{ 6 } \\ \Rightarrow A=\dfrac { 3 }{ 2\pi } $$
Question 5
1 / -0

A wave is represented by the equation $$y=10 \sin 2\pi(100t-0.02x)+10 \sin 2\pi(100+0.02x)$$. The maximum amplitude and loop length are respectively

A
$$20$$ units and $$30$$ units

B
$$20$$ units and $$25$$ units

C
$$30$$ units and $$20$$ units

D
$$25$$ units and $$20$$ units

Solution

$$x= 10\sin 2\pi \Big(100t-0.02x\Big)+10\sin 2\pi \Big(100t+0.02x\Big)$$.
$$\Rightarrow 10 [\sin A+\sin B]$$,
where
$$B= 2\pi \Big(100t-0.02x\Big)$$ and
$$A= 2\pi \Big(100t+0.02x\Big)$$.
Thus,
$$\Rightarrow 10 [2\sin \dfrac{A+B}{2}\sin \dfrac{A-B}{2}]$$
$$\Rightarrow 20\sin (2\pi 100t) \sin (2\pi 0.02x)$$.
Comparing the above equation with standard standing wave equation, we get
amplitude $$= 20$$ and wave vector$$k=\dfrac{2\pi}{\lambda}=2\pi \times 0.02 \Rightarrow \lambda = 50 $$.
Therefore, the loop length = $$\dfrac{\lambda}{2}= 25$$
Question 6
1 / -0

Equations of motion in the same direction is given by
$$y_1 = A\sin(\omega t - kx)$$, $$\space y_2 = A\sin(\omega t - kx - \theta)$$. The amplitude of the medium particle will be

A
$$\sqrt2\ A\cos\theta$$

B
$$2A\cos\theta$$

C
$$\sqrt2A\cos\dfrac{\theta}{2}$$

D
$$2A\cos\dfrac{\theta}{2}$$

Solution

Amplitude of the resultant wave is
$$A_R=\sqrt{A^2_1+A^2_2+2A_1A_2\cos(\theta)}$$.
Here, $$A_1=A_2=A$$, so
$$A_R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)$$.
Question 7
1 / -0

Two waves of equal amplitude $$x_0$$ and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is

A
$$0$$

B
$$x_0$$

C
$$2x_0$$

D
Between $$0$$ and $$2x_0$$

Solution

The amplitude of the resultant wave is
$$x'_0=\sqrt{x^2_{01}+x^2_{02}+2x_{01}x_{02}\cos(\phi)}$$, here the angle between the two propagating waves is $$\phi$$.
$$x'_{0}$$ is maximum when $$\phi = 0^o$$ and it becomes $$x'_{0}=x_{01}+x_{02}=2x_0$$ for $$x_{01}=x_{02}=x_0$$.
$$x'_{0}$$ is minimum when $$\phi = 180^o$$ and it becomes $$x'_{0}=x_{01}-x_{02}=0$$ for $$x_{01}=x_{02}=x_0$$.
The amplitude of the resultant wave is between 0 and 2$$x_0$$.
Question 8
1 / -0

A wave of frequency $$400\space Hz$$ has a velocity of $$320\space ms^{-1}$$. The distance between the particles differing in phase by $$90^{\small\circ}$$ is

A
$$80\space cm$$

B
$$60\space cm$$

C
$$40\space cm$$

D
$$20\space cm$$

Solution

Let,
Frequency of wave is, $$f = 400 Hz$$
Velocity of wave is, $$v = 320 ms^{-1}$$
The phase difference is, $$\phi = 90^\circ = \dfrac{\pi}{2}$$
The wavelength of wave is, $$\lambda$$
$$\lambda = \dfrac{v}{f}$$
$$\lambda = \dfrac{320}{400}$$
$$\lambda = 0.8 m$$
Now, the phase difference between two particles in a wave is
$$\dfrac{\phi}{2\pi}= \dfrac{x}{\lambda}$$
where, x is the distance between two particles in a wave.
$$x = \dfrac{\lambda \phi}{2 \pi}$$
$$x = \dfrac{0.8 (\dfrac{\pi}{2})}{2 \pi}$$
$$x = \dfrac{0.8}{4} = 0.2 m$$
$$x = 20 cm$$
Question 9
1 / -0

The resultant amplitude due to superposition of two waves $$y_{1} = 5 \sin (wt-kx)$$ and $$y_{2}=-5 \cos (wt-kx-150^{o})$$

A
$$5$$

B
$$5\sqrt{3}$$

C
$$5\sqrt{2-\sqrt{3}}$$

D
$$5\sqrt{2+\sqrt{3}}$$

E
$$5\sqrt{3-\sqrt{2}}$$

Solution

Given :
$${ y }_{ 1 }=5\sin(wt+kx)\\ { y }_{ 2 }=5\cos(wt+kx+{ 150 }^{ \circ })\\ y={ y }_{ 1 }+{ y }_{ 2 }$$
where $$y$$ is the resultant wave
The amplitude of this resultant wave is given by the formula:
$$ A = \sqrt { { (A }_{ 1 })^ 2+({ A }_{ 2 })^ 2+2{ A }_{ 1 }{ A }_{ 2 }\cos\phi } $$
Here;
$${ A }_{ 1 }=5$$
$${ A }_{ 2 }=5$$
$$\phi ={ 150 }^{ \circ }$$
on substituting the values on the given formula:
$$ A=\sqrt { (5)^ 2+(5)^ 2+2\times 5\times 5\times \cos{ 150 }^{ \circ } } $$
$$ =\sqrt { 25+25-2\times 25\times \frac { 1 }{ 2 } } .................(\cos{ 150 }^{ \circ }=-\dfrac { 1 }{ 2 } )$$
Hence A=5 (option A is correct)
Question 10
1 / -0

Two waves of same frequency but amplitudes equal to $$a$$ and $$2a$$ travelling in the same direction superimpose out of phase. The resultant amplitude will be

A
$$\sqrt{a^2+2a^2}$$

B
$$3a$$

C
$$2a$$

D
$$a$$

Solution

The amplitude of the resultant wave is
$$x'_0=\sqrt{a^2+ (2a)^2+2 a \times 2a\cos(\phi)}$$, here the angle between the two propagating waves is $$\phi=\pi$$.
Hence $$x'_0=\sqrt{a^2+ (2a)^2+2 a \times 2a\cos(\pi)}=\sqrt{a^2+ (2a)^2-2 a \times 2a}=a$$

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Waves Test - 27

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Waves Test - 27

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Question 1

Both, sound and light waves in air are longitudinal.

Both, sound and light waves in air are transverse.

Sound waves in air are transverse and light waves are longitudinal.

Sound waves in air are longitudinal and light waves are transverse.

Solution

Question 2

2 m/s

3 m/s

4 m/s

6 m/s

Solution

Question 3

rarefaction

crest

trough

compression

Solution

Question 4

$$\displaystyle\frac{3}{2\pi}m$$

$$\displaystyle\frac{3}{4\pi}m$$

$$\displaystyle\frac{6}{\pi}m$$

$$\displaystyle\frac{3}{8\pi}$$

Solution

Question 5

$$20$$ units and $$30$$ units

$$20$$ units and $$25$$ units

$$30$$ units and $$20$$ units

$$25$$ units and $$20$$ units

Solution

Question 6

$$\sqrt2\ A\cos\theta$$

$$2A\cos\theta$$

$$\sqrt2A\cos\dfrac{\theta}{2}$$

$$2A\cos\dfrac{\theta}{2}$$

Solution

Question 7

$$0$$

$$x_0$$

$$2x_0$$

Between $$0$$ and $$2x_0$$

Solution

Question 8

$$80\space cm$$

$$60\space cm$$

$$40\space cm$$

$$20\space cm$$

Solution

Question 9

$$5$$

$$5\sqrt{3}$$

$$5\sqrt{2-\sqrt{3}}$$

$$5\sqrt{2+\sqrt{3}}$$

$$5\sqrt{3-\sqrt{2}}$$

Solution

Question 10

$$\sqrt{a^2+2a^2}$$

$$3a$$

$$2a$$

$$a$$

Solution

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