Waves Test

Result

Waves Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

A nylon guitar string has a linear density of $$7.20\ g/m$$ and is under tension of $$150\ N$$. The fixed supports are distance $$D= 90.0\ cm$$ apart. The string is oscillating in the standing wave pattern shown in figure.Calculate the
(iii) The frequency of the traveling waves whose superposition gives this standing wave.

A
$$\frac{1000}{\sqrt[3]{3}}Hz$$

B
$$\frac{1250}{\sqrt[3]{3}}Hz$$

C
$$\frac{1500}{\sqrt[3]{3}}Hz$$

D
$$\frac{1750}{\sqrt[3]{3}}Hz$$
Question 2
1 / -0

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

B
$$y = -1 \sin(ax + bt)$$

C
$$y = -8.1 \sin(ax - bt)$$

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

Solution
Question 3
1 / -0

Two waves of equal frequencies have their amplitudes in the ratio of 3 : 5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

A
16:1

B
15:1

C
1:16

D
1:15

Solution
Question 4
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A particle executes SHM with a time period of $$16$$s. At time $$t=2s$$, the particle crosses the mean position while at $$t=4s$$, its velocity is $$4ms^{-1}$$. The amplitude of motion in metre is?

A
$$\sqrt{2}\pi$$

B
$$16\sqrt{2}\pi$$

C
$$32\sqrt{2}/\pi$$

D
$$4/\pi$$

Solution

$$x=\alpha sin\left( \dfrac { 2\pi }{ T } t+\phi \right) $$

$$t=2s,x=0,T=16s$$

$$0=\alpha sin\left( \dfrac { \pi }{ 4 } +\phi \right) $$

$$\phi =-\dfrac { \pi }{ 4 } $$

equ of shm is

$$x=asin\left( \dfrac { 2\pi }{ T } t-\frac { \pi }{ 4 } \right) $$

$$t=4s,v=4m/s$$

$$v=\dfrac { dx }{ dt } =a\times \dfrac { 2\pi }{ t } cos\left( \dfrac { 2\pi }{ T } t-\dfrac { \pi }{ 4 } \right) $$

$$4=\alpha \times \dfrac { 2\pi }{ 16 } cos\left( \dfrac { \pi }{ 2 } -\dfrac { \pi }{ 4 } \right) $$

$$4=\alpha \times \dfrac { 2\pi }{ 16 } \times \dfrac { \pi }{ 8 } \times \dfrac { 1 }{ \sqrt { 2 } } $$

$$\alpha =\dfrac { 32\sqrt { 2 } }{ \pi } $$
Question 5
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If two waves, each of intensity $${I}_{0}$$, having the same frequency but differing by a constant phase angle of $${60}^{o}$$, superpose at a certain point in space, then the intensity of resultant wave is:

A
$$2{I}_{0}$$

B
$$\sqrt{3}{I}_{0}$$

C
$$3{I}_{0}$$

D
$$4{I}_{0}$$

Solution
Question 6
1 / -0

A particle is executing SHM of amplitude $$A$$, about the mean position $$x = 0$$. Which of the following cannot be a possible phase difference between the positions of the particle at $$x = +A/2$$ and $$x = -A/\sqrt {2}$$.

A
$$75^{\circ}$$

B
$$165^{\circ}$$

C
$$135^{\circ}$$

D
$$195^{\circ}$$
Question 7
1 / -0

Two particles execute $$S.H.M.$$ along the same line at the same frequency. They move in opposite direction at the mean position. The phase difference will be:

A
$$1\pi$$

B
$$2\pi/3$$

C
$$\pi$$

D
$$\pi/2$$

Solution
Question 8
1 / -0

Two waves are represented by $$x_1 = A \sin \left( \omega t + \dfrac{\pi}{6} \right) $$ and $$x_2 = A \cos \omega t $$ then the phase difference between them is :

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

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

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

D
$$ \pi $$

Solution

The general equation of the waves given is

$${x_1} = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)$$

$${x_2} = A\cos \left( {\omega t} \right)$$

Then $${x_1}$$ can be written as
$${x_1} = A\cos \left( {\dfrac{\pi }{2} - \omega t - \dfrac{\pi }{6}} \right)$$
On comparing and solving we get the phase difference is$$\dfrac{\pi }{3}$$
Question 9
1 / -0

Equation of a progressive wave is given by
$$y = 0.2 \cos \pi \Bigg \lgroup 0.04t + .02x - \frac{\pi}{6} \Bigg \rgroup$$
The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of $$\pi$$/2

A
4 cm

B
8 cm

C
25 cm

D
12.5 cm

Solution
Question 10
1 / -0

Two waves are propagating to the point p along a straight line produced by two sources A and B of ahead by $$\pi /3$$ than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter

A
$$2a$$

B
$$a\sqrt{3}$$

C
$$a\sqrt{2}$$

D
$$a$$

Solution