Waves Test

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Waves Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Potential energy of a string depends on

A
Wave velocity

B
Amplitude of the wave

C
Extent of stretching of the string

D
None of the above

Solution

Potential energy of a string depends on the extent of stretching.

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

A wave travelling on a string as shown in figure gets reflected at an open boundary. Then,

A
Phase reversal takes place at the boundary

B
Wavelength changes at the boundary

C
Phase reversal dosent take place

D
Speed of the reflected ray takes place

Solution

Phase reversal dosent take place , since reflection takes place at an open boundary. The reflected wave will look like as shown in the figure

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

The equation of an incident wave travelling along +X direction is given by $$y= A sin (2t-5x)$$. This wave gets reflected at a rigid boundary. The equation of the reflected wave is

A
$$y= A sin (2t-5x)$$

B
$$y= A sin (2t-5x +\pi)$$

C
$$y= A sin (2t+5x +\pi)$$

D
$$y= A sin (2t-5x +\pi/2)$$

Solution

A reflected wave will be $$\pi$$ radians out of phase with the incident wave. Apart from this, the wave will also get reflected along the negative x axis. Thus, the sign of x also changes.

The equation of the reflected wave is thus $$y= A sin (2t+5x +\pi)$$
The correct option is (c)
Question 4
1 / -0

The equation of a progressive wave is given by $$y=10 sin (5t-x)$$. The wave gets reflected from a open boundary. The equation of the reflected wave is

A
$$y=10 sin (5t-x)$$

B
$$y=-10 sin (5t-x)$$

C
$$y=10 sin (5t+x)$$

D
$$y=10 sin (5t+x+\pi)$$

Solution

In reflection of a wave at open boundaries, the phase reversal dosent take place

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

The total energy per unit length for a travelling wave in a string of mass density $$\mu$$ , whose wave function is $$A(x,t) = f(x \pm vt)$$ is given by:

A
$$E_tot = \sqrt(\mu/2) (\dfrac{dA}{dt})^2$$

B
$$E_tot = (\mu/2) (\dfrac{dA}{dt})^2$$

C
$$E_tot = (\mu/2)^2 (\dfrac{dA}{dt})^2$$

D
$$E_tot = (2\mu) (\dfrac{dA}{dt})^2$$
Question 6
1 / -0

Two waves represented by $$y_1=10\sin(2000\pi t)$$ and $$y_2=10\sin (2000 \pi t+\pi /2)$$ are superimposed at any point at a particular instant. The resultant amplitude is?

A
$$10$$ units

B
$$20$$ units

C
$$14.1$$ units

D
Zero

Solution

The resultant amplitude A of two waves of amplitudes $$a_1$$ and $$a_2$$ at a phase difference of $$\phi$$ is $$(a^2_1+a^2_2+2a_1a_2\cos\phi)^{1/2}$$. Substituting $$a_1=10, a_2=10$$ and $$\phi =90^o$$,
we get $$A=14.1$$
Hence (C) is correct.
Question 7
1 / -0

Kinetic energy per unit length for a particle in a standing wave is zero at:

A
nodes

B
antinodes

C
mid-way between a node and an antinode

D
None of the above

Solution

Particle at antinodes is momentarily at rest and hence has zero kinetic energy. Its speed comes down to zero at this point and all energy is stored in the form of potential energy.
Question 8
1 / -0

A travelling wave has an equation of the form $$A(x,t)=f(x+vt)$$. The relation connecting positional derivative with time derivative of the function is:

A
$$\dfrac{dA}{dt}=\pm v^2 \dfrac {dA}{dx}$$

B
$$\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$$

C
$$\dfrac{dA}{dt}=\pm \sqrt(v) \dfrac {dA}{dx}$$

D
$$\dfrac{dA}{dt}=(2 \pi v/\lambda) \dfrac {dA}{dx}$$

Solution

Positional derivative and time derivative of a function f is $$\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$$

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

The angular frequency of a particle in a progressive wave in an elastic medium is $$100\pi\ { rads }^{ -1 }$$ and it is moving with a velocity of $$200{ms}^{-1}$$. The phase difference between two particles separated by a distance of $$20m$$ is:

A
$$31.4\ rad$$

B
$$\pi\ rad$$

C
$$\dfrac{3\pi}{4}\ rad$$

D
$$36\ rad$$

Solution
Question 10
1 / -0

A string vibrates according to equation $$y=\sin { \cfrac { \pi x }{ 3 } } \cos { 40\pi t } $$. The potential energy of the string will be zero at times

A
$$\cfrac{1}{20}s$$ and $$\cfrac{1}{40}s$$

B
$$\cfrac{1}{40}s$$ and $$\cfrac{1}{80}s$$

C
$$\cfrac{1}{80}s$$ and $$\cfrac{3}{80}s$$

D
$$\cfrac{1}{40}s$$ and $$\cfrac{3}{40}s$$

Solution

$$y=sin\dfrac { \pi x }{ 3 } \cos40\pi t$$
The potential energy depends on the displacement of particles from the equilibrium position. So P.E is zero only when displacement is $$0$$.
So let $$y=0$$
$$\sin\left( \dfrac { \pi x }{ 3 } \right) \cos\left( 40\pi t \right) =0$$
$$\Rightarrow \quad \cos\left( 40\pi t \right) =0$$
$$40\pi t=\left( 2n+1 \right) \dfrac { \pi }{ 2 } $$ (where $$n=$$ integer)
$$80t=2n+1$$
$$\left[ t=\dfrac { 2n+1 }{ 80 } sec \right] $$
$$\Rightarrow \quad t=\dfrac { 1 }{ 80 } ,\dfrac { 3 }{ 80 } ,\dfrac { 5 }{ 80 } ,............seconds$$

Hence Option (C) is correct.

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Re-Attempt Weekly Quiz Competition

Waves Test - 48

Result

Waves Test - 48

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SHARING IS CARING

Question 1

Wave velocity

Amplitude of the wave

Extent of stretching of the string

None of the above

Solution

Question 2

Phase reversal takes place at the boundary

Wavelength changes at the boundary

Phase reversal dosent take place

Speed of the reflected ray takes place

Solution

Question 3

$$y= A sin (2t-5x)$$

$$y= A sin (2t-5x +\pi)$$

$$y= A sin (2t+5x +\pi)$$

$$y= A sin (2t-5x +\pi/2)$$

Solution

Question 4

$$y=10 sin (5t-x)$$

$$y=-10 sin (5t-x)$$

$$y=10 sin (5t+x)$$

$$y=10 sin (5t+x+\pi)$$

Solution

Question 5

$$E_tot = \sqrt(\mu/2) (\dfrac{dA}{dt})^2$$

$$E_tot = (\mu/2) (\dfrac{dA}{dt})^2$$

$$E_tot = (\mu/2)^2 (\dfrac{dA}{dt})^2$$

$$E_tot = (2\mu) (\dfrac{dA}{dt})^2$$

Question 6

$$10$$ units

$$20$$ units

$$14.1$$ units

Zero

Solution

Question 7

nodes

antinodes

mid-way between a node and an antinode

None of the above

Solution

Question 8

$$\dfrac{dA}{dt}=\pm v^2 \dfrac {dA}{dx}$$

$$\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$$

$$\dfrac{dA}{dt}=\pm \sqrt(v) \dfrac {dA}{dx}$$

$$\dfrac{dA}{dt}=(2 \pi v/\lambda) \dfrac {dA}{dx}$$

Solution

Question 9

$$31.4\ rad$$

$$\pi\ rad$$

$$\dfrac{3\pi}{4}\ rad$$

$$36\ rad$$

Solution

Question 10

$$\cfrac{1}{20}s$$ and $$\cfrac{1}{40}s$$

$$\cfrac{1}{40}s$$ and $$\cfrac{1}{80}s$$

$$\cfrac{1}{80}s$$ and $$\cfrac{3}{80}s$$

$$\cfrac{1}{40}s$$ and $$\cfrac{3}{40}s$$

Solution

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