Waves Test

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Waves Test - 82

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Weekly Quiz Competition

Question 1
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A harmonic wave has been set up on a very long string which travels along the length of the string. The wave has frequency of $$50 \,Hz,$$ amplitude $$1 \,cm$$ and wavelength $$0.5 \,m.$$ For the above-described wave.
$$Statement \ I :$$ Time taken by the wave to travel a distance of $$8 \,m$$ along the length of the string is $$0.32 \,s.$$
$$Statement \ II :$$ Time taken by a point on the string to travel a distance of $$8 \,m,$$ once the wave has reached at that point and sets it into motion is $$0.32 \,s.$$

A
Both the statement are correct

B
Statement I is correct but Statement II is incorrect

C
Statement I is incorrect but Statement II is correct

D
Both the statement are incorrect.

Solution

The wave is travelling along the length of a string, while particles constituting the string are oscillation in a direction perpendicular to the length to string. In one time period (cycle) , the wave moves forward by one wavelength while the particle on string travels a distance of 4 times the amplitude.
Here, $$T = 1/f = 0.02 \,s$$
Wave speed, $$v = f\lambda = 25 \,m/s$$
Time taken by wave to travel a distance of 8 m.
$$t_1 = 8/25 \,s = 0.32 \,s.$$
Time taken by particle on string to travel a distance of 8 m,
$$t_2 = \dfrac {8 \times T}{4 \,times \,amplitude} = \dfrac {8}{4 \times 0.01} \times 0.02 = 4 \,s$$
Question 2
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Directions For Questions

A $$2\ kg$$ block hangs without vibrating at the bottom end of a spring with a force constant of $$800\ N/m$$. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upwards acceleration of $$10\ m/s^{2}$$ when the acceleration suddenly ceases at time $$t=0$$ and the car moves upward with constant speed $$(g=10\ m/s^{2})$$.

...view full instructions

The initial phase angle observed by a rider in the elevator,taking downward direction to be positive and positive extreme position to have $$\pi/2$$ phase constant, is equal to:

A
zero

B
$$\pi/2\ rad$$

C
$$\pi/ rad$$

D
$$3\pi/2\ rad$$

Solution

As elevator was rising up with $$a=10\,m/s^2$$ pseudo force was acting on block.
$$F=ma+mg$$ in downward direction.
Spring Force =$$Kx$$ in upward direction
previous mean position be $$x_o$$
$$F_{net}=0$$
$$ma+mg=Kx_o$$

$$x_o=\dfrac{ma+mg}{K}$$

$$x_o=\dfrac{2\times 10+2\times 10}{800}$$

$$x_o=0.050 m=5 cm$$

This is extreme position now.
now, pseudo force stop acting , new mean position be $$x'$$

$$F_{net}=0$$
$$mg=Kx'$$

$$x'=\dfrac{mg}{K}$$

$$x'=\dfrac{2\times 10}{800}$$
$$x'=0.025 m=2.5 cm$$

so initially block moves from $$x=5 cm$$ to $$x=2.5 cm$$,
i.e block moves upwards
and it is given that downward direction is positive.
which means block is moving in negative direction,
so phase difference would be $$-\pi/2$$ i.e $$3\pi/2$$
Question 3
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Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite direction each time their displacement us half their amplitude. Their phase difference is :

A
$$\dfrac{5\pi}{6}$$

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

C
$$\dfrac{\pi}{6}$$

D
$$\dfrac{2\pi}{3}$$

Solution
Question 4
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A standing wave arises on a string when two waves of equal amplitude, frequency, and wavelength traveling in opposite directions superimpose. If the frequency of two-component wäves is doubled, then the frequency of oscillation of the standing waves

A
gets doubled

B
gets halved

C
remains unchanged

D
changes but not by a factor of 2 or $$ 1 / 2 $$

Solution
Question 5
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Directions For Questions

A longitudinal standing wave y = a cos kx cos$$ \omega t$$ is maintained in a homogenous medium of density $$\rho$$. Here $$\omega$$ is the angular speed and k, the wavenumber, and a is the amplitude of the standing wave. This standing wave exists all over a given region of space.

...view full instructions

The space density of the potential energy $$PE = E_p(x, t)$$ at a point (x, t) in the space is

A
$$E_p = \dfrac{\rho a^2 \omega^2}{2}$$

B
$$E_p = \dfrac{\rho a^2 \omega^2}{2} cos^2\space kx\space sin^2 \omega t$$

C
$$E_p = \dfrac{\rho a^2 \omega^2}{2} sin^2\space kx\space cos^2 \omega t$$

D
$$E_p = \dfrac{\rho a^2 \omega^2}{2} sin^2\space kx\space sin^2 \omega t$$

Solution
Question 6
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Directions For Questions

Consider a standing wave formed on a string. It results due to the superposition of two waves travelling in opposite directions. The waves are travelling along the length of the string in the x- direction and displacement of elements on the string are along the y- direction. Individual equations of the two waves can be expressed as
$$Y_1 = 6(cm) sin [ 5 (rad/cm) x - 4 (rad/s) t]$$
$$Y_2 = 6(cm) sin [5 (rad/cm) x + 4 ( rad/s) t]$$

Here x and y are is cm.

...view full instructions

Figure shows the standing wave pattern at t = 0 due to superposition of waves given by $$y_1\space and\space y_2$$. N is anode and A an antinode. At this instant say t = 0, instantaneous velocity of points on the string

A
is different fro different points

B
is zero for all points

C
changes with position of the point

D
is constant but not equal to zero for all points

Solution

$$y = y_1 + y_2 = (12 sin 5x) cos 4t$$

Maximum value of y-position in SHM of an element of the string that is located at an antinode = $$\pm 12 cm\space (sin 5x = pm 1)$$
For the position nodes amplitude should be xoro.
So, Sin 5x = 0 $$\Rightarrow 5x = n\pi$$
$$ x = \dfrac{n\pi}{5}$$
Where n = 0, 1, 2, 3,......
Value of amplitude at x = 1.8 cm
A = 12 sin (5 X 1.8) = 4.9 cm
At any instant say t = 0, instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero.
Question 7
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The following equations represent progressive transverse waves
$$z_1 = A \cos (\omega t - kx)$$
$$z_2 = A \cos (\omega t + kx)$$
$$z_3 = A \cos (\omega t + ky)$$
$$z_4 = A \cos (2\omega t - 2ky)$$
A stationary wave will be formed by superposing

A
$$z_1$$ and $$z_2$$

B
$$z_1$$ and $$z_4$$

C
$$z_2$$ and $$z_3$$

D
$$z_3$$ and $$z_4$$

Solution

a. The direction of wave must be opposite and frequencies will be same then by susperposition, standing wave formation takes place.
Question 8
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Directions For Questions

Consider a standing wave formed on a string. It results due to the superposition of two waves travelling in opposite directions. The waves are travelling along the length of the string in the x- direction and displacement of elements on the string are along the y- direction. Individual equations of the two waves can be expressed as
$$Y_1 = 6(cm) sin [ 5 (rad/cm) x - 4 (rad/s) t]$$
$$Y_2 = 6(cm) sin [5 (rad/cm) x + 4 ( rad/s) t]$$

Here x and y are is cm.

...view full instructions

Amplitude of simple harmonic motion of a point on the string that is located at x = 1.8 cm will be

A
3.3 cm

B
6.7 cm

C
4.9 cm

D
2.6 cm

Solution

$$y = y_1 + y_2 = (12 sin 5x) cos 4t$$

Maximum value of y-position in SHM of an element of the string that is located at an antinode = $$\pm 12 cm\space (sin 5x = pm 1)$$
For the position nodes amplitude should be xoro.
So, Sin 5x = 0 $$\Rightarrow 5x = n\pi$$
$$ x = \dfrac{n\pi}{5}$$
Where n = 0, 1, 2, 3,......
Value of amplitude at x = 1.8 cm
A = 12 sin (5 X 1.8) = 4.9 cm
At any instant say t = 0, instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero.
Question 9
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Directions For Questions

A longitudinal standing wave y = a cos kx cos$$ \omega t$$ is maintained in a homogenous medium of density $$\rho$$. Here $$\omega$$ is the angular speed and k, the wavenumber, and a is the amplitude of the standing wave. This standing wave exists all over a given region of space.

...view full instructions

The space density of the kinetic energy, $$KE = E_k(x, t)$$ at the point (x, t) is given by

A
$$E_k = \dfrac{\rho a^2 \omega^2}{2} cos^2\space kx\space cos^2 \omega t$$

B
$$E_k = \dfrac{\rho a^2 \omega^2}{2} sin^2\space kx\space cos^2 \omega t$$

C
$$E_k = \dfrac{\rho a^2 \omega^2}{2} $$

D
$$E_k = \dfrac{\rho a^2 \omega^2}{2} cos^2\space kx\space sin^2 \omega t$$

Solution
Question 10
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Directions For Questions

Consider a standing wave formed on a string. It results due to the superposition of two waves travelling in opposite directions. The waves are travelling along the length of the string in the x- direction and displacement of elements on the string are along the y- direction. Individual equations of the two waves can be expressed as
$$Y_1 = 6(cm) sin [ 5 (rad/cm) x - 4 (rad/s) t]$$
$$Y_2 = 6(cm) sin [5 (rad/cm) x + 4 ( rad/s) t]$$

Here x and y are is cm.

...view full instructions

Maximum value of the y-position coordinate in the simple harmonic motion of an element of the string that is located at an antinode will be

A
$$\pm 6 cm$$

B
$$\pm 8 cm$$

C
$$\pm 12 cm$$

D
$$\pm 3 cm$$

Solution