Oscillations Test

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Oscillations Test - 24

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Question 1
1 / -0

Ratio of kinetic energy at mean position to potential energy at $$A/2$$ of a particle performing SHM.

A
$$2:1$$

B
$$4:1$$

C
$$8:1$$

D
$$1:1$$

Solution

Kinetic energy $$K=\displaystyle\frac{1}{2}m\omega^2(A^2-y^2)$$
At mean position $$y=0$$
$$K=\displaystyle\frac{1}{2}m\omega^2(A^2)$$ .......$$(i)$$
Potential energy $$U=\displaystyle\frac{1}{2}m\omega^2y^2$$
$$U$$ at $$y=\displaystyle\frac{A}{2}$$
$$U=\displaystyle\frac{1}{2}\frac{mA^2}{4}\omega^2$$ ........$$(ii)$$
Dividing Eq $$(i)$$ by Eq $$(ii)$$, we get
$$\displaystyle\frac{KE}{U}=\frac{4}{1}$$
Question 2
1 / -0

The frequency of oscillator of the springs as shown in figure will be

A
$$\displaystyle \frac {1}{2\pi} \sqrt {\frac {(k_1+k_2)m}{k_1k_2}}$$

B
$$\displaystyle \frac {1}{2\pi} \sqrt {\frac {k_1k_2}{(k_1+k_2)m}}$$

C
$$\displaystyle \frac {1}{2\pi}\sqrt {\frac {k}{m}}$$

D
$$\displaystyle 2\pi\sqrt {\frac {k}{m}}$$

Solution

$$\displaystyle n=\frac {1}{2\pi}\sqrt{\frac {k}{m}}$$
As springs are in parallel, total spring constant $$k$$ of system of spring
$$\displaystyle \frac {1}{k} = \frac {1}{k_1} + \frac {1}{k_2} = \frac {k_1+k_2}{k_1k_2} k = \frac {k_1k_2}{(k_1+k_2)}$$

$$\displaystyle n=\frac {1}{2\pi} \sqrt {\frac {k_1k_2}{(k_1+k_2)m}}$$
Question 3
1 / -0

All oscillatory motions are ______

A
periodic

B
linear

C
rotatory

D
curvilinear

Solution

All oscillatory is periodic, as in all oscillatory motion particle repeats its position after a certain interval of time.
Question 4
1 / -0

Which is not a necessary assumption for simple pendulum

A
in-extensible, light and flexible string.

B
heavy but small sized sphere (bob).

C
angular displacement from the mean position is very less.

D
volume of the bob is negligible.

Solution

All the other three option are important assumption used in while deriving the time period of Simple Harmonic Motion.
Question 5
1 / -0

A pendulum suspended from the ceiling of a train has a time period T when the train is at rest. When the train is accelerating with a uniform acceleration, the time period will

A
increase

B
decrease

C
become infinite

D
remain unaffected

Solution

When a train is moving with acceleration, the pendulum is exerted by a pseudo acceleration opposite to the direction of the train. As a result, there will be an increase in effective acceleration on the pendulum. The time period will decrease as the time period is inversely proportional to the square root of effective acceleration.
Question 6
1 / -0

If a pendulum takes $$4 sec$$ to swing in each direction, find the length of the pendulum.

A
4.2 m

B
3.97 m

C
5.2 m

D
15.9 m

Solution

Given: The pendulum takes 4sec to swing in each direction, which means it swings in one side of the pendulum for 4 sec.
So, half time period, $$\frac{T}{2}= 4\ sec$$
Time period of simple pendulum, $$T= 2\times 4=8 s$$
$$T =2\pi\sqrt{\dfrac{l}{g}}$$
$$\Rightarrow 8 =2\pi\sqrt{\dfrac{l}{10}}$$
$$\Rightarrow l =\frac{64\times 10}{4\times \pi^2}$$

$$\Rightarrow l\approx 15.9\ m$$
Question 7
1 / -0

For the block under SHM shown in the figure, which of the following quantities is directly proportional to the maximum speed of the block?

A
Amplitude

B
Frequency

C
Period

D
Position of block

E
Total mechanical energy of the block

Solution

The maximum potential energy of a spring is $$U_{max}=\dfrac{1}{2}kA^2$$ where k be the force constant and A be the amplitude of the oscillation.
When the block passes through the equilibrium position, the all potential energy is converted into kinetic energy and at this point the speed will be maximum.
Thus, $$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv_{max}^2$$
so, $$A \propto v_{max} $$
Question 8
1 / -0

A particle is executing simple harmonic motion with an amplitude $$A$$ and time period $$T$$. The displacement of the particles after $$2 T$$ period from its initial position is

A
$$A$$

B
$$4A$$

C
$$8A$$

D
Zero

Solution

It is the least interval of time after which the periodic motion of a body repeats itself. Therefore, displacement will be zero.
Question 9
1 / -0

Find out the Time period of simple harmonic oscillator vibrating with frequency $$2.5$$ Hz and an amplitude of $$0.05$$m:

A
$$0.4 sec$$

B
$$0.2 sec$$

C
$$8 sec$$

D
$$20 sec$$

E
$$50 sec$$

Solution

We know that frequency $$f=\dfrac{1}{T}$$
So, time period , $$T=\dfrac{1}{f}=\dfrac{1}{2.5}=0.4 sec$$
The option A will be correct.
Question 10
1 / -0

A block of mass $$m$$ attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where :

A
the speed is the maximum

B
the potential energy is the minimum

C
the speed is the minimum

D
the restoring force is the minimum

E
the kinetic energy is the maximum

Solution

The magnitude of acceleration of the simple harmonic motion is $$a=F/m=kx/m$$
So, the acceleration is maximum when the displacement $$x$$ from equilibrium will be maximum. Thus, speed will be minimum and also the potential energy will also be maximum and kinetic energy will be minimum at the end points of the oscillation.

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Oscillations Test - 24

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Oscillations Test - 24

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SHARING IS CARING

Question 1

$$2:1$$

$$4:1$$

$$8:1$$

$$1:1$$

Solution

Question 2

$$\displaystyle \frac {1}{2\pi} \sqrt {\frac {(k_1+k_2)m}{k_1k_2}}$$

$$\displaystyle \frac {1}{2\pi} \sqrt {\frac {k_1k_2}{(k_1+k_2)m}}$$

$$\displaystyle \frac {1}{2\pi}\sqrt {\frac {k}{m}}$$

$$\displaystyle 2\pi\sqrt {\frac {k}{m}}$$

Solution

Question 3

periodic

linear

rotatory

curvilinear

Solution

Question 4

in-extensible, light and flexible string.

heavy but small sized sphere (bob).

angular displacement from the mean position is very less.

volume of the bob is negligible.

Solution

Question 5

increase

decrease

become infinite

remain unaffected

Solution

Question 6

4.2 m

3.97 m

5.2 m

15.9 m

Solution

Question 7

Amplitude

Frequency

Period

Position of block

Total mechanical energy of the block

Solution

Question 8

$$A$$

$$4A$$

$$8A$$

Zero

Solution

Question 9

$$0.4 sec$$

$$0.2 sec$$

$$8 sec$$

$$20 sec$$

$$50 sec$$

Solution

Question 10

the speed is the maximum

the potential energy is the minimum

the speed is the minimum

the restoring force is the minimum

the kinetic energy is the maximum

Solution

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