Waves Test

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Waves Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

A man is travelling towards east at 20ms$$^{-1}$$ and sound waves are moving towards east at 340ms$$^{-1}$$. He finds that $$500$$ waves cross him in $$2$$ seconds, the wavelength of the sound wave received by the man is :

A
$$1.28 m$$

B
$$2.28 m$$

C
$$3.28 m$$

D
$$4.28 m$$

Solution

Speed of sound relate to man is $$=340-20$$ $$=320\ m/s$$
$$f=\dfrac{number\ of\ waves\ crossing }{time}$$$$=\dfrac{500}{2}$$$$=250\ Hz$$
$$f=\dfrac{u}{\lambda }$$
$$\lambda=\dfrac{u}{f}$$$$=\dfrac{320}{250}$$$$=1.28m$$
Question 2
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The equation $$y=4+2\sin(6t-3x)$$ represents a wave motion with amplitude of

A
$$6 units$$

B
$$2 units$$

C
$$20 units$$

D
$$8 units$$

Solution

A wave can be represented by:
$$y=C+a\sin(\omega t \pm kx)$$, where a is the amplitude of the wave. Comparing with the equation given in the question, $$a=2$$.
Question 3
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The equation of a progressive wave is $$ y = 0.5 \sin \ 2\pi (\dfrac{t}{0.004} - \dfrac{x}{50})$$, where $$x$$ is in $$cm$$. The phase difference between two points separated by a distance $$5 \ cm$$ at any instant is :

A
$$\dfrac \pi 5 $$

B
$$\dfrac \pi 3 $$

C
$$\dfrac \pi 2 $$

D
$$\pi $$

Solution

$$Wavelength(\lambda)=\dfrac{2\pi}{k}= \dfrac{2\pi \times 50}{2\pi}=50\: cm$$
We know that path difference of $$\lambda$$ gives rise to phase difference of $$2\pi$$. So path difference of $$5\: cm$$ gives rise to phase difference of $$\dfrac{2\pi}{\lambda} \times 5=\dfrac{\pi}{5}$$
Question 4
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The equation of a plane progressive wave is $$y = 0.9 \sin 4\pi [t-\displaystyle \dfrac{x}{2}]$$ , when it is reflected at a rigid support, its amplitude becomes $$2/3$$ of its previous value. The equation
of the reflected wave is :

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

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

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

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

Solution

$$y=0.9\ \sin\ 4\ \pi (t-\dfrac{x}{2})$$
Reflected wave equation will be opposite in direction and of amplitude $$\dfrac{2}{3}\times 0.9=0.6$$ and amplitude will be negative of incident wave.
So, $$y_{reflected}=-0.6\ \sin\ 4\pi \left ( t+\dfrac{x}{2} \right ).$$
Question 5
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In a stationary wave that forms as a result of reflection of waves from an obstacle, the ratio of the amplitude at an antinode to the amplitude at node is $$6$$. The percentage of energy transmitted is

A
$$25\%$$

B
$$49\%$$

C
$$58\%$$

D
$$72\%$$

Solution

the ratio of the amplitude at an antinode to the amplitude at node is 6, so
$$\dfrac{a_i+a_r}{a_i-a_r}=6$$
or, $$\dfrac{a_r}{a_i}=5/7$$
Now, the energy is proportional to the square of the amplitudes.
So, $$\dfrac{E_r}{E_i}=25/49=51\%$$
So the rest will be the transmitted energy, i.e. $$49\%$$.
Question 6
1 / -0

Statement-1 : Where two vibrating tuning forks having frequencies $$256 Hz$$ and $$512 Hz$$ are held near each other, beats cannot be heard.
Statement-2 : The principle of superposition is valid only if the frequencies of the oscillations are nearly equal.

A
Both the statements are true and statement-2 is the correct explanation of statement-1

B
Both the statements are true but statement-2 is not the correct explanation of statement-1

C
Statement-1 is true and statement-2 is false

D
Statement-1 is false and statement-2 is true

Solution

For beats to be heard the two frequencies should be close to each other. When two frequencies are $$f$$ and $$2f$$, the beat frequency is $$2f-f=f$$ which is the original frequency.

Thus no beats are heard. Principle of super position is valid for all frequencies.
Question 7
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Two waves of equal amplitude $$x_{o}$$ and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is :

A
$$0$$

B
$$x_{o}$$

C
$$2x_{o}$$

D
Between $$0$$ and $$2x_{o}$$

Solution

Two waves of equal amplitude $$x_o$$ and equal frequency travel in the same direction in a medium.
$$\therefore$$ Amplitude of the resultant wave is given by,
$$R=x_1+x_2$$
$$R$$ is minimum when both are zero.
$$R$$ is maximum when both are $$x_o$$.
$$\therefore R_{max}=x_o+x_o=2x_o$$
$$\therefore $$ Amplitude of the resultant wave is given between $$0$$ and $$2x_o$$
Question 8
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A wave pulse on a string has the dimension shown in figure. The wave speed is $$\nu= 1 \ cm/ s$$. If point O is a free end, the shape of wave at time $$t = 3s$$ is :

A

B

C

D

Solution

In three seconds, the distance travelled by wave=$$1cm/s\times 3s=3cm$$
Hence the center of the wave, the crest, reaches the point O' in the time.
The amplitude of the wave becomes $$1cm+1cm=2cm$$
Hence correct answer is option D.
Question 9
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Two sound waves, each of amplitude $$A$$ and frequency $$\omega$$, superpose at a point with a phase difference of $$\displaystyle \dfrac{\pi}{2}$$. The amplitude and the frequency of the resultant wave are, respectively

A
$$A^2,\omega$$

B
$$2A,\omega$$

C
$$A,\omega$$

D
$$\sqrt{2}A,\omega$$

Solution

Let the two waves be $$A\ \sin\ wt,\ A\ \cos\ wt$$
$$\therefore y=y_{1}+y_{2}=\ A\ \sin\ wt+A\ \cos\ wt$$
$$=\sqrt{2}A\left ( \dfrac{1}{\sqrt{2}}\sin\ wt+\ \cos\ wt\dfrac{1}{\sqrt{2}} \right )$$
$$=\sqrt{2}A\ \ \ \sin \left ( wt+\dfrac{\pi}{4} \right )$$
amplitude is $$\sqrt{2}A$$, frequency is same, i.e. $$w$$
Question 10
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A wave represented by $$y=100\sin(ax+bt)$$ is reflected from a dense plane at the origin. If $$36\%$$ of energy is lost and rest of the energy is reflected, then the equation of the reflected wave will be

A
$$ y=-8.1\sin ({\it ax-bt} )$$

B
$$y=8.1\sin(ax+bt)$$

C
$$ y=-80\sin (ax-bt )$$

D
$$ y=-10\sin (ax-bt)$$

Solution

Given, $$36\%$$ of energy is lost, energy of wave reflected back is $$64\%$$
$$I \alpha A^{2} \Rightarrow \ \ \ \ A_{2}=\dfrac{8}{10}\times A_{1}$$
$$\therefore \dfrac{8}{10}\times 100=80$$
reflected wave will be in opposite direction as it's reflected from a dense plane.
$$\Rightarrow y=-80\ \sin\ (ax-bt)$$