Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 3

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The shortest distance travelled by a particle (performing S.H.M.) from mean position in $2$ seconds is equal to $\frac{\sqrt{3}}{2}$ of its amplitude. Find its period.

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Important Questions on Oscillations

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 4

The motion of a particle varies with time according to the relation
y = a sin ωt + a cos ωt. Then

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 5

A particle executing SHM of amplitude

4 cm

and

T = 4 s

. The time taken by it to move from positive extreme position to half the amplitude is:

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 6

In

SHM,

the acceleration of a particle at its mean position is

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 7

A child swinging on a swing in sitting position stands up. Then the time period of the swing will

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 8

The displacement of a particle in simple harmonic motion in one time period is,

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 9

What is the phase difference between two simple harmonic motions represented by

x_{1} = A \sin (ω t + \frac{π}{6})

and

x_{2} = A \cos (ω t)

?

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 1

The circular motion of a particle with constant speed is

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Achieve Indian Airforce Agniveer Vayu Physics Practice Book>Chapter 14 - Oscillations>Exercise>Q 2

A particle executes simple harmonic motion between

x = - A a n d x = + A

. The time taken for it to go from 0 to

A / 2 i s T_{1}

and to go from

A / 2 t o A i s T_{2}

. Then