Oscillations Test

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Oscillations Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

For a simple pendulum, relation between time period $$T$$, length of pendulum $$l$$ and acceleration due to gravity $$g$$ is given by :

A
$$T=2 \pi \dfrac{l}{\sqrt{g}}$$

B
$$T=2 \pi \sqrt{\dfrac{l}{g}} $$

C
$$T=2 \pi \dfrac{l}{g}$$

D
None of these

Solution

Time period of simple pendulum is given by $$T = 2\pi\sqrt{\dfrac{l}{g}}$$
where $$l$$ is the length of pendulum and $$g$$ is the acceleration due to gravity.
Question 2
1 / -0

When a simple pendulum is taken from the Equator to the Pole, its period of oscillation will :

A
increase

B
decrease

C
remains same

D
can't say

Solution

Time period of simple pendulum is given by $$T =2\pi\sqrt{\dfrac{l}{g}}$$
where $$l$$ is the length of the pendulum and $$g$$ is the acceleration due to gravity.
We know that $$g_{pole}> g_{equator}$$
$$\implies \ T_{pole}<T_{equator}$$
Thus time period of simple pendulum decreases when it is taken from equator to pole.
Question 3
1 / -0

A tunnel is dug along the diameter of the earth. A mass $$m$$ is dropped into it. How much time does it take to cross the earth?

A
$$169.2\ minutes$$

B
$$84.6\ minutes$$

C
$$21.2\ minutes$$

D
$$42.3\ minutes$$

Solution

The mass will execute SHM in the tunnel.
Time period of oscillation $$T = 2\pi\sqrt{\dfrac{R}{g}}$$
where $$R = 6.4\times 10^6 \ m$$ and $$g = 9.8 \ m/s^2$$
$$\implies \ T = 2\pi\sqrt{\dfrac{6.4\times 10^6}{9.8}} \ s = 5075 \ s$$
So we get $$T = \dfrac{5075}{60} = 84.6 \ minutes$$
Time taken to cross the tunnel $$t = \dfrac{T}{2} = \dfrac{84.6}{2} = 42.3 \ minutes$$
Question 4
1 / -0

One end of a rod of length L is fixed to a point on the circumference of a wheel of radius R. The other end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure. The wheel is rotating with a constant angular velocity $$\omega$$ about O. Taking $$\displaystyle T=\frac{2\pi}{P\omega}$$ the motion of the rod is?

A
Simple harmonic with a period of T

B
Simple harmonic with a period of $$T/2$$

C
Not simple harmonic but periodic with a period of T

D
Not simple harmonic but periodic with a period of $$T/2$$

Solution

In simple harmonic motion, oscillations occur about an equilibrium position where net force is zero. Also, at extreme positions, the object is stationary. Hence, the motion is not SHM.

The rod moves periodically because after each time period $$T$$, rod returns in the original position.
Question 5
1 / -0

A mass $$m$$ is suspended from the two coiled spring which have the same length in unstreched condition. Two springs are connected in parallel. The force constant for springs are $$K_{1}$$ and $$K_{2}$$. The time period of the suspended mass will be

A
$$T = 2\pi \sqrt {\left (\dfrac {m}{K_{1} + K_{2}}\right )}$$

B
$$T = 2\pi \sqrt {\left (\dfrac {m}{K_{1} - K_{2}}\right )}$$

C
$$T = 2\pi \sqrt {\left (\dfrac {m(K_{1} + K_{2})}{K_{1}K_{2}}\right )}$$

D
$$T = 2\pi \sqrt {\left (\dfrac {m(K_{1}K_{2})}{K_{1} + K_{2}}\right )}$$

Solution

K is equivalent spring constant of combination of $${ k }_{ 1 }$$ and $${ k }_{ 2 }$$
$$k={ k }_{ 1 }={ k }_{ 2 }$$
$$T=2\pi \sqrt { \cfrac { m }{ k } } =2\pi \sqrt { \cfrac { m }{ { k }_{ 1 }+{ k }_{ 2 } } } $$
Question 6
1 / -0

The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $${E}_{1}$$ and $${E}_{2}$$ are the total mechanical energies respectively. Then

A
$${ E }_{ 1 }=\sqrt { 2 } { E }_{ 2 }$$

B
$${ E }_{ 1 }=2{ E }_{ 2 }$$

C
$${ E }_{ 1 }=4{ E }_{ 2 }$$

D
$${ E }_{ 1 }=16{ E }_{ 2 }$$

Solution

For, a S.H.M the total mechanical energy $$E\propto { A }^{ 2 }$$ ($$A$$=amplitude)
In case of small circle amplitude $$a$$ and energy $${ E }_{ 2 }$$ and in case of large circle amplitude is $$2a$$ and energy $${ E }_{ 1 }$$.
So, $$\cfrac { { E }_{ 2 } }{ { E }_{ 1 } } =\cfrac { { (a) }^{ 2 } }{ { (2a) }^{ 2 } } $$
$$\Rightarrow { E }_{ 1 }=4{ E }_{ 2 }$$
$$*$$N.B: In case of circular motion the radius of circle donates the amplitude of S.H.M when it is projected along its diameter.
Question 7
1 / -0

A simple harmonic motion has an amplitude $$A$$ and time period $$T$$. The time required by it to travel $$x = A$$ to $$x = \dfrac {A}{2}$$ is:

A
$$T / 6$$

B
$$T / 4$$

C
$$T / 3$$

D
$$T / 2$$

Solution

$$x=A\cos\omega t$$
$$t=0,x=A$$
At $$x=\dfrac{A}{2} $$,
$$\dfrac{A}{2}=A\cos\omega t$$
$$\omega t=\dfrac{\pi}{3}$$
$$\dfrac{2\pi}{T}t=\dfrac{\pi}{3}$$
$$t=\dfrac{T}{6}$$ where T is time period
Question 8
1 / -0

A spring has a certain mass suspended from it and its period for vertical oscillation is $$T$$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now

A
$$\dfrac {T}{2}$$

B
$$\dfrac {T}{\sqrt {2}}$$

C
$$\sqrt {2}T$$

D
$$2T$$

Solution

We can treat spring of length $$l$$ and spring constant $$k$$ as series combination of two spring of two spring of constant $$k$$ and length $$\cfrac { l }{ 2 } $$. So,$$k=\cfrac { k'k' }{ k'+k' } $$
$$\Rightarrow k'=2k$$
In first case, $$T=2\pi \sqrt { \cfrac { m }{ k } } $$
and in second case, $$T'=2\pi \sqrt { \cfrac { m }{ k' } } =2\pi \sqrt { \cfrac { m }{ 2k } } =\cfrac { 1 }{ \sqrt { 2 } } (2\pi \sqrt { \cfrac { m }{ k } } )$$
$$\Rightarrow T'=\cfrac { T }{ \sqrt { 2 } } $$
Question 9
1 / -0

When the cylinder is slightly depressed and released,it oscillates.Let there be a mean position.Find time period

A
$$\pi { \left[ \dfrac { \left( n+1 \right) L }{ g\left( n-1 \right) } \right] }^{ \dfrac { 1 }{ 2 } }$$

B
$$\pi { \left[ \dfrac { { n }^{ 2 }L }{ g{ \left( 1-n \right) }^{ 2 } } \right] }^{ \dfrac { 1 }{ 2 } }$$

C
$$\pi { \left[ \dfrac { { n }L }{ g\left( { n }^{ 2 }-1 \right) } \right] }^{ \dfrac { 1 }{ 2 } }$$

D
$$\pi { \left[ \dfrac { { n }L }{ g{ \left( 1-n \right) }^{ 2 } } \right] }^{ \dfrac { 1 }{ 2 } }$$
Question 10
1 / -0

A particle performs S.H.M of amplitude $$A$$ along a straight line. When it is at distance $$\cfrac { \sqrt { 3 } }{ 2 } A$$ from mean position, its kinetic energy gets increased by an amount $$\cfrac { 1 }{ 2 } m{ \omega }^{ 2 }{ A }^{ 2 }$$ due to an impulsive force. Then its new amplitude becomes

A
$$\cfrac { \sqrt { 5 } }{ 2 } A$$

B
$$\cfrac { \sqrt { 3 } }{ 2 } A$$

C
$$\sqrt { 2 } A$$

D
$$\sqrt { 5 } A$$

Solution

Given,

A S.H.M have amplitude A.

So, its total Initial Kinetic Energy of the system is $$\dfrac{1}{2}m{{\omega }^{2}}A$$

Energy given to the system is $$\dfrac{1}{2}m{{\omega }^{2}}A$$. This energy gets added to system energy.

Total final Kinetic energy is = Total initial kinetic energy + added kinetic energy

$$ \dfrac{1}{2}m{{\omega }^{2}}A{{'}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}+\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}} $$

$$ A'=A\sqrt{2} $$

Hence, New amplitude is $$A\sqrt{2}$$

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 29

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Oscillations Test - 29

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SHARING IS CARING

Question 1

$$T=2 \pi \dfrac{l}{\sqrt{g}}$$

$$T=2 \pi \sqrt{\dfrac{l}{g}} $$

$$T=2 \pi \dfrac{l}{g}$$

None of these

Solution

Question 2

increase

decrease

remains same

can't say

Solution

Question 3

$$169.2\ minutes$$

$$84.6\ minutes$$

$$21.2\ minutes$$

$$42.3\ minutes$$

Solution

Question 4

Simple harmonic with a period of T

Simple harmonic with a period of $$T/2$$

Not simple harmonic but periodic with a period of T

Not simple harmonic but periodic with a period of $$T/2$$

Solution

Question 5

$$T = 2\pi \sqrt {\left (\dfrac {m}{K_{1} + K_{2}}\right )}$$

$$T = 2\pi \sqrt {\left (\dfrac {m}{K_{1} - K_{2}}\right )}$$

$$T = 2\pi \sqrt {\left (\dfrac {m(K_{1} + K_{2})}{K_{1}K_{2}}\right )}$$

$$T = 2\pi \sqrt {\left (\dfrac {m(K_{1}K_{2})}{K_{1} + K_{2}}\right )}$$

Solution

Question 6

$${ E }_{ 1 }=\sqrt { 2 } { E }_{ 2 }$$

$${ E }_{ 1 }=2{ E }_{ 2 }$$

$${ E }_{ 1 }=4{ E }_{ 2 }$$

$${ E }_{ 1 }=16{ E }_{ 2 }$$

Solution

Question 7

$$T / 6$$

$$T / 4$$

$$T / 3$$

$$T / 2$$

Solution

Question 8

$$\dfrac {T}{2}$$

$$\dfrac {T}{\sqrt {2}}$$

$$\sqrt {2}T$$

$$2T$$

Solution

Question 9

$$\pi { \left[ \dfrac { \left( n+1 \right) L }{ g\left( n-1 \right) } \right] }^{ \dfrac { 1 }{ 2 } }$$

$$\pi { \left[ \dfrac { { n }^{ 2 }L }{ g{ \left( 1-n \right) }^{ 2 } } \right] }^{ \dfrac { 1 }{ 2 } }$$

$$\pi { \left[ \dfrac { { n }L }{ g\left( { n }^{ 2 }-1 \right) } \right] }^{ \dfrac { 1 }{ 2 } }$$

$$\pi { \left[ \dfrac { { n }L }{ g{ \left( 1-n \right) }^{ 2 } } \right] }^{ \dfrac { 1 }{ 2 } }$$

Question 10

$$\cfrac { \sqrt { 5 } }{ 2 } A$$

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

$$\sqrt { 2 } A$$

$$\sqrt { 5 } A$$

Solution

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