Oscillations Test

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Oscillations Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

A particle of mass $$3$$kg, attached to a spring with force constant $$48$$ N/m execute simple harmonic motion on a frictionless horizontal surface. The time period of oscillation of the particle, in seconds, is?

A
$$\dfrac{\pi}{4}$$

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

C
$$2\pi$$

D
$$8\pi$$

E
$$\dfrac{\pi}{8}$$

Solution

The time period of mass,
$$T=2\pi\sqrt{\dfrac{m}{k}}$$
Given, $$m=3$$kg
$$k=48$$N/m
$$T=2\pi\sqrt{\dfrac{3}{48}}=2\pi\sqrt{\dfrac{1}{16}}$$
$$=2\pi\times \dfrac{1}{4}=\dfrac{\pi}{2}$$.
Question 2
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The displacement of a particle varies with time, according to the equation $$y = a \sin \omega t + b \cos \omega t$$

A
The motion is oscillatory but not SHM.

B
The motion is SHM. with amplitude (a + b)

C
The motion is SHM. with amplitude $$(a^{2} + b^{2})$$

D
The motion is SHM. with amplitude $$\sqrt{a^{2} + b^{2}}$$

Solution

(d) : $$y = a \sin \omega t + b \cos \omega t$$

The amplitude of motion is given by

$$A = \sqrt{a^{2} + b^{2}}$$

Hence, verifies the option (d).
Question 3
1 / -0

The diagram shows a ball hanging on a string. The ball swings from point $$W$$ to point $$Z$$ and back to point $$W$$.
Which statement about the ball is correct?

A
The kinetic energy of the ball is greatest at point $$W$$

B
The kinetic energy of the ball is greatest at point $$X$$

C
The kinetic energy of the ball is greatest at point $$Y$$

D
The kinetic energy of the ball is the same at all points of the swing

Solution

It is a case of simple pendulum .
we know that
by conservation of energy
$$KE+PE=$$ constant.
At point Y , PE is min.
So, the kinetic energy at mean position is maximum in this case.
Since ,Y is the mean position , so, the kinetic energy is max. at point $$Y$$.
Question 4
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The velocity of a particle performing simple harmonic motion, when it passes through its mean position is

A
Infinity

B
Zero

C
Minimum

D
Maximum

Solution

Equation of SHM particle
$$Y=a\sin \omega t\\V=a\omega \sin\omega t$$
$$V_{max}=a\omega$$
Question 5
1 / -0

The length of simple pendulum is increased by $$1\%$$. Its period will

A
Increase by $$1\%$$

B
Increase by $$0.5\%$$

C
Decrease by $$0.5\%$$

D
Increase by $$2\%$$

Solution

$$T\propto \sqrt {l}$$

$$\Rightarrow \dfrac {\Delta T}{T}=\dfrac {1}{2}\dfrac {\Delta l}{l}=\dfrac {1}{2}\times 1\% =0.5\%$$
Question 6
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If a simple pendulum is taken to place where $$g$$ decreases by $$2\%$$ then the time period

A
Decreases by $$1\%$$

B
Increases by $$2\%$$

C
Increase by $$2\%$$

D
Increases by $$1\%$$

Solution

Time period for simple pendulam,
$$T=2\pi \sqrt{\dfrac{l}{g}}$$
$$\Rightarrow T\propto \dfrac{1}{\sqrt{g}}$$

$$\Rightarrow \dfrac{\Delta T}{T}\times 100=-\dfrac{1}{2}\left(\dfrac{\Delta g}{g}\right)\times 100=-\dfrac{1}{2}(-2\%)=1\%$$
Question 7
1 / -0

A simple pendulum is vibrating in an evacuated chamber, it will oscillate with

A
Increasing amplitude

B
Constant amplitude

C
Decreasing amplitude

D
First $$(c)$$ then $$(a)$$

Solution

As it is clear that in vacuum, the bob will not experienced any frictional force. Hence therefore shall be no dissipation therefore, it will oscillation with constant amplitude.
Question 8
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The amplitude of a particle performing S.H.M. is $$A$$. The displacement at which its velocity will be half of the maximum velocity is

A
$$A/2$$

B
$$A/3$$

C
$$ \sqrt{3}A/2 $$

D
$$2A/\sqrt{3} $$

Solution

$$x = A sin \omega t$$
$$V=\dfrac{dx}{dt}=A\omega cos \omega t$$
$$V_{max} =A \omega$$
$$V=\dfrac{V_{max}}{2}= A \omega cos \omega t$$
$$\Rightarrow \dfrac{A \omega }{2}=A \omega cos \omega t$$
$$\Rightarrow cos \omega t=\dfrac{1}{2}$$
$$\Rightarrow \omega t =60^{0}$$
$$x=A sin \omega t$$
$$=A sin 60^{0}$$
$$=A \dfrac{\sqrt{3}}{2}$$
so, amplitude is $$\dfrac{\sqrt{3}A}{2}$$
Question 9
1 / -0

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

A
$$ \dfrac{\pi}{2} $$

B
$$ 2\pi $$

C
$$ \pi $$

D
$$ \dfrac{\pi}{4} $$

Solution

In SHM
$$a=-\omega^2x$$
or
$$a= \dfrac{-K}{m}x$$
so, from graph
$$-\dfrac{K}{m}= -1$$ $$ ( \because slope\ is\ -1)$$
$$\dfrac{K}{m}= 1$$
Time period $$= 2\pi \sqrt{\dfrac{m}{K}}$$
$$= 2\pi \sqrt{\dfrac{1}{1}}$$
$$= 2\pi $$
Question 10
1 / -0

The equation of the displacement of two particles making SHM are represented by
$$ y_{1} $$ = a sin $$ \left ( \omega t +\phi \right ) $$ & $$ y_{2} $$ = a cos $$ \left ( \omega t \right ) $$.
The phase difference of the velocities of the two particles is :

A
$$ \frac{\pi }{2} +\phi $$

B
$$ - \phi $$

C
$$ \phi $$

D
$$ \phi -\frac{\pi }{2} $$

Solution

$$y_{1}=a sin (\omega t+\phi  )$$
$$y_{2}=a   cos (\omega t)$$
$$y_{1}=a   \omega cos (\omega t+\phi)=a \omega sin (\omega t+\phi +{\pi}/_{2})$$
$$y_{2}=-a   \omega sin (\omega t)= a \omega sin (\pi + \omega t)$$
therefore phase difference$$ = \omega t+\phi +{\pi}/_{2}-(\pi+\omega t)$$
$$=\phi -{\pi}/_{2}$$

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 14

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Oscillations Test - 14

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Question 1

$$\dfrac{\pi}{4}$$

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

$$2\pi$$

$$8\pi$$

$$\dfrac{\pi}{8}$$

Solution

Question 2

The motion is oscillatory but not SHM.

The motion is SHM. with amplitude (a + b)

The motion is SHM. with amplitude $$(a^{2} + b^{2})$$

The motion is SHM. with amplitude $$\sqrt{a^{2} + b^{2}}$$

Solution

Question 3

The kinetic energy of the ball is greatest at point $$W$$

The kinetic energy of the ball is greatest at point $$X$$

The kinetic energy of the ball is greatest at point $$Y$$

The kinetic energy of the ball is the same at all points of the swing

Solution

Question 4

Infinity

Zero

Minimum

Maximum

Solution

Question 5

Increase by $$1\%$$

Increase by $$0.5\%$$

Decrease by $$0.5\%$$

Increase by $$2\%$$

Solution

Question 6

Decreases by $$1\%$$

Increases by $$2\%$$

Increase by $$2\%$$

Increases by $$1\%$$

Solution

Question 7

Increasing amplitude

Constant amplitude

Decreasing amplitude

First $$(c)$$ then $$(a)$$

Solution

Question 8

$$A/2$$

$$A/3$$

$$ \sqrt{3}A/2 $$

$$2A/\sqrt{3} $$

Solution

Question 9

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

$$ 2\pi $$

$$ \pi $$

$$ \dfrac{\pi}{4} $$

Solution

Question 10

$$ \frac{\pi }{2} +\phi $$

$$ - \phi $$

$$ \phi $$

$$ \phi -\frac{\pi }{2} $$

Solution

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