Oscillations Test

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Oscillations Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

A particle is executing SHM with time period T Starting from mean position, time taken by it to complete $$\dfrac { 5 }{ 8 } $$ oscillations is:

A
$$\dfrac { T }{ 12 } $$

B
$$\dfrac { T }{ 6 } $$

C
$$\dfrac { 5T }{ 12 } $$

D
$$\dfrac { 7T }{ 12 } $$

Solution
Question 2
1 / -0

The equation of motion of a particle executing SHM is $$\left( \dfrac { d ^ { 2 } x } { d t ^ { 2 } } \right) + k x = 0$$ The time period of the particle will be

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

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

C
$$2 \pi k$$

D
$$2 \pi \sqrt { k }$$

Solution
Question 3
1 / -0

A particle performs $$SHM$$ on x-axis with amplitude $$A$$ and time period $$T$$. The time taken by the particle to a distance $$A/5$$ starting from rest is:

A
$$\dfrac { T }{ 20 }$$

B
$$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 4 }{ 5 }\right)$$

C
$$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 1 }{ 5 }\right)$$

D
$$\dfrac { T }{ 2\pi } \sin^{-1} \left(\dfrac { 4 }{ 5 }\right)$$

Solution

Particle is starting from rest, i.e, from one of its extreme position.
As particle moves a distance $$A/5$$, we can represent it on a circle as shown.
$$\cos \theta=\dfrac{4A/5}{A}=\dfrac{4}{5}\Rightarrow \theta=\cos^{-1}\left( \dfrac{4}{5}\right)$$
$$\omega t=\cos^{-1}\left( \dfrac{4}{5}\right)\Rightarrow t=\dfrac{1}{\omega}\cos^{-1}\left( \dfrac{4}{5}\right)=\dfrac{T}{2\pi}\cos^{-1}\left( \dfrac{4}{5}\right)$$
Question 4
1 / -0

The figure shows the displacement time graph of a particle executing S.H.M.
If the time period of oscillation is $$2 s$$ the equation of motion of its SHM

A
$$
x = 10 \sin ( \pi t + \pi / 3 )
$$

B
$$
x = 10 \sin \pi t
$$

C
$$
x = 10 \sin ( \pi t + \pi / 6 )
$$

D
$$
x = 10 \sin (2 \pi t + \pi / 6 )
$$

Solution
Question 5
1 / -0

A coin is placed on a horizontal platform, which undergoes horizontal simple harmonic motion about a mean position $$O$$.The coin does not slip on the platform. The force of friction acting on the coin is $$F$$.

A
$$F$$ is always directed towards $$O$$

B
$$F$$ is directed towards $$O$$ when the coin is moving away from $$O$$, and away from $$O$$ when the coin moves towards $$O$$

C
$$F=0$$ when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.

D
$$F$$ is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.

Solution

Coin does not slip on the platform, its mean that Coin will also execute SHM with same frequency with the platform so, Its acceleration will always act towards the O, hence force will be always towards the O.
Question 6
1 / -0

a simple harmonic has an amplitude A and time period T. Find the time required by it to travel directly from $$x=0$$ to $$\, x=\frac { A}{ \sqrt { 2} } $$

A
T

B
2T

C
T/2

D
T/8

Solution
Question 7
1 / -0

When the displacement is half of the amplitude, then what fraction of total energy of a simple harmonic oscillator is kinetic:-

A
$$\dfrac {3}{4}th$$

B
$$\dfrac {2}{7}th$$

C
$$\dfrac {5}{7}th$$

D
$$\dfrac {2}{8}th$$

Solution
Question 8
1 / -0

The displacement $$x$$ (in metres) of a particle performing simple harmonic motion is related to time $$t$$ (in seconds as} $$x = 0.05 \cos \left( 4 \pi t + \frac { \pi } { 4 } \right)$$. The frequency of the motion will be

A
$$0.5$$$$\mathrm { Hz }$$

B
$$1.0$$$$\mathrm { Hz }$$

C
$$1.5$$$$\mathrm { Hz }$$

D
$$2.0$$$$\mathrm { Hz }$$

Solution

Comparing the given equation with the standard equation of SHM i.e. $$y=a\ sin(\omega t+\phi)$$

$$y=0.05 sin\left(\dfrac{\pi}{2}-(4\pi t+\dfrac{\pi}{4})\right)$$

$$y=0.05\ sin(\dfrac{\pi}{4}-4\pi t)$$

So, the angular frequency of the oscillation will be:
$$\omega=4\pi$$

So, frequency of oscillation will be:
$$f=\dfrac{\omega}{2\pi}$$

$$f=\dfrac{4\pi}{2\pi}=2\ Hz$$
Question 9
1 / -0

Restoring force on the bob of a simple pendulum of mass 100 gm when its amplitude is $${1^\circ}$$ :

A
$$1.7 N$$

B
$$0.17 N$$

C
$$0.017 N$$

D
$$0.034 N$$
Question 10
1 / -0

A simple harmonic motion is represented by $$y = 5 ( \sin 3 \pi t + \sqrt { 3 } \cos 3 \pi t ) \mathrm { cm }$$ The amplitude and time period of the motion are :

A
$$10 \mathrm { cm } , \dfrac { 2 } { 3 } \mathrm { s }$$

B
$$5 \mathrm { cm } , \dfrac { 2 } { 3 } \mathrm { s }$$

C
$$10 \mathrm { cm } , \dfrac { 3 } { 2 } \mathrm { s }$$

D
$$5 \mathrm { cm } , \dfrac { 3 } { 2 } \mathrm { s }$$

Solution

$$y=5(\sin3\pi t+\sqrt{3}\cos 3\pi t)$$
$$y=5\times 2\left(\dfrac{1}{2}\sin 3\pi t+\dfrac{\sqrt{3}}{2}\cos 3\pi t\right)$$
$$y=10\left(\cos \dfrac{\pi}{3}\sin 3\pi t-\sin \dfrac{\pi}{3}\cos 3\pi t\right)$$
$$y=10\sin\left(3\pi t+\dfrac{\pi}{3}\right)$$
Comparing it with $$y=A\sin(\omega t+\phi)$$
$$\Rightarrow A=10$$ i.e., amplitude is $$10$$cm
$$\Rightarrow T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{3\pi}=\dfrac{2}{3}$$ sec. i.e, time period is $$\dfrac{2}{3}$$ sec.