Waves Test

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Waves Test - 75

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Weekly Quiz Competition

Question 1
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Maximum acceleration of an object in simple harmonic motion is 24 $$m/s^2$$ and maximum velocity is 16 m/sec. The amplitude of object is

A
$$\dfrac { 32 } { 3 } \mathrm { m }$$

B
$$16\ m$$

C
$$\dfrac { 2 } { 3 } \mathrm { m }$$

D
$$\dfrac { 3 } { 2 } \mathrm { m }$$

Solution
Question 2
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Directions For Questions

A block of mass $$M$$ is suspended from on end of a light spring as shown. The origin $$O$$ is considered at distance equal to the natural length of the spring from the ceiling and vertical downward direction as positive $$y-$$ axis. When the system is in equilibrium, a bullet of mass $$m/3$$ moving in a vertical upward direction with velocity $$v_{0}$$ strikes the block and embeds into it. As a result, the block (with a bullet embedded into it) moves up and starts oscillating.
Based on the given information answer the following question:

...view full instructions

The time taken by block-bullet system to move from $$y = \dfrac { m g } { k }$$ (initial equilibrium position) to $$y = 0$$ (natural length of spring) is ($$A$$ represents the amplitude of motion).

A
$$\sqrt { \dfrac { 4 m } { 3 k } } \left[ \cos ^ { - 1 } \left( \dfrac { m g } { 3 k A } \right) - \cos ^ { - 1 } \left( \dfrac { 4 m g } { 3 k A } \right) \right]$$

B
$$\sqrt { \dfrac { 3 k } { 4 m } } \left[ \cos ^ { - 1 } \left( \dfrac { m g } { 3 k A } \right) - \cos ^ { - 1 } \left( \dfrac { 4 m g } { 3 k A } \right) \right]$$

C
$$\sqrt { \dfrac { 4 \mathrm { m } } { 6 k } } \left[ \sin ^ { - 1 } \left( \dfrac { 4 \mathrm { mg } } { 3 k A } \right) - \sin ^ { - 1 } \left( \dfrac { \mathrm { mg } } { 3 k A } \right) \right]$$

D
$$None\ of\ the\ above$$

Solution

Initially in equilibrium let the elongation is spring be $$y_{0}$$ then $$mg=ky_{0}$$
$$y_{0}=\dfrac{mg}{k}$$
As the bullet strikes the block with velocity $$v_{0}$$ and gets embedded into it, the velocity of the combined mass can be computed by using the principle of moment conservation.
$$\dfrac{m}{3}v_{0}=\dfrac{4m}{3}v\Rightarrow v=\dfrac{v_{0}}{4}$$
Let new mean position is at distance $$y$$ from the origin, then
$$ky=\dfrac{4m}{3}g\Rightarrow y=\dfrac{4 mg}{3k}$$

Now , the block executes $$SHM$$ about mean position defined by $$y=4 mg/3k$$ with time period $$T=2\pi \sqrt{4m/3k}$$. At $$t=0$$, the combined mass is at a displacement of $$(y-y_{0})$$ from mean position and is moving with velocity $$v$$, then by using $$v=\omega \sqrt{A^{2}-x^{2}}$$, we can find the amplitude of motion.

$$\left(\dfrac{v_{0}}{4}\right)^{2}=\dfrac{3k}{4m}[A^{2}-(y-y_{0})^{2}]=\dfrac{3k}{4m}\left[A^{2}-\left(\dfrac{mg}{3k}\right)^{2}\right]$$

$$\Rightarrow A=\sqrt{\dfrac{mv_{0}^{2}}{12k}+\left(\dfrac{mg}{3k}\right)^{2}}$$
To compute the time taken by the combined mass from $$y=mg/k$$ to $$y=0$$, we can either go for equation method or circuit motion projection method.

Required time, $$t=\dfrac{\theta}{\omega}=\dfrac{\alpha-\beta}{\omega}$$

$$\cos a=\dfrac{y-y_{0}}{A}=\left(\dfrac{\dfrac{4mg}{3k}-\dfrac{4mg}{k}}{A}\right)=\dfrac{mg}{3kA}$$

$$\cos\beta=\dfrac{y}{A}=\dfrac{4mg}{3kA}$$

So, $$t=\dfrac{\cos\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)}{\omega}$$
$$t=\sqrt{\dfrac{4m}{3k}}\left[\cos^{-1}\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)\right]$$
Question 3
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A person observe two points on a string as a travelling wave passes them. The points are at $$x _ { 1 } = 0$$ and $$x_2 = 1m$$. The transverse motions of the two points are found to be as follows: $$y _ { 1 } = 0.2 \sin 3 \pi t$$
$$y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 )$$ What is the frequency in $$Hz$$?

A
$$1.5 Hz$$

B
$$3 Hz$$

C
$$4.5 Hz$$

D
$$1 Hz$$

Solution
Question 4
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A wave pulse is given by the equation $$y=f(x,t)=A exp(-B(x-vt)^{2})$$. Given $$A=1.0m.B=1.0m^{-2}$$ and $$v=+2.0m/s$$. which of the following graph shows the correct wave profile at the instant $$t=1s$$?

A

B

C

D

Solution
Question 5
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A particle starts from the origin, goes along $$X-$$ axis to the point $$(20\ m,\ 0)$$ and then returns along the same line to the point $$(-20m,\ 0)$$. The distance and displacement of the particle during the trip are

A
$$40m$$, $$0$$

B
$$40m$$, $$20m$$

C
$$40m$$, $$-20m$$

D
$$60m$$, $$20m$$
Question 6
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A particle executes simple harmonic motion with a time period of 16 s. At time t=2s, the particle crosses the mean position. Its velocity is $$4 ms^{-1}$$ when t=4s. The amplitude of motion is

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

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

C
$$24 \sqrt{2}\, \pi m$$

D
$$\dfrac{32\sqrt{2}}{\pi}m$$

E
$$\dfrac{4}{\pi}m$$

Solution
Question 7
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A particle executes SHM with a time period of 16 second .at time t is equal 2 second the particle crosses the mean position while at t is equal to 4 second its velocity is 4 m per second 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
Question 8
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The distance between consecutive maxima and minima is given by

A
$$ \lambda / 2 $$

B
$$ 2 \lambda $$

C
$$ \lambda $$

D
$$ \lambda / 4 $$
Question 9
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The amplitude of a wave represented by displacement equation :
$$ y=\frac { 1 }{ \sqrt { a } } sin\omega t\pm \frac { 1 }{ \sqrt { b } } cos\omega t $$ will be :

A
$$\sqrt{ \dfrac { { a+b } }{ ab } }$$

B
$$ \dfrac { \sqrt { a } +\sqrt { b } }{ ab } $$

C
$$ \dfrac { \sqrt { a } -\sqrt { b } }{ ab } $$

D
$$ \dfrac { a+b }{ ab } $$

Solution
Question 10
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A body is performing linear SHM, If the displacement, acceleration and the corresponding velocity of the body are y, 'a' and v respectievly, which of the following graphs is/are correct?

A

B

C

D

Solution