Question

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

Answer 1

Given: The amplitude of simple harmonic motion is 5 cm and the period of oscillation of the disc is 0.2 s.

The angular velocity of the body is given as,

ω= 2π T

Where, Tis the time period of oscillation of the body.

The acceleration of the body is given as,

A=− ω 2 y(1)

The velocity of the particle is given as,

V=ω r 2 − y 2 (2)

a)

The displacement of the particle is 5 cm.

The acceleration of the body is given as,

A=− ω 2 y

By substituting the given values in the above equation, we get

A=− ( 10π ) 2 ( 0.05 ) =−5 π 2  m/ s 2

The velocity of the body is given as,

V=ω r 2 − y 2

By substituting the given values in the above equation, we get

V=( 10π ) ( 0.05 ) 2 − ( 0.05 ) 2 =0 m/s

Thus, the acceleration of the body is −5 π 2  m/ s 2 and the velocity of the body is 0 m/s .

b)

The displacement of the particle is 3 cm.

The acceleration of the body is given as,

A=− ω 2 y

By substituting the given values in the above equation, we get

A=− ( 10π ) 2 ( 0.03 ) =−3 π 2  m/ s 2

The velocity of the body is given as,

V=ω r 2 − y 2

By substituting the given values in the above equation, we get

V=( 10π ) ( 0.05 ) 2 − ( 0.03 ) 2 =0.4π m/s

Thus, the acceleration of the body is −3 π 2  m/ s 2 and the velocity of the body is 0.4π m/s .

c)

The displacement of the particle is 0 cm.

The acceleration of the body is given as,

A=− ω 2 y

By substituting the values in the above equation, we get

A=− ( 10π ) 2 ( 0 ) =0 m/ s 2

The velocity of the body is given as,

V=ω r 2 − y 2

By substituting the given values in the above equation, we get

V=( 10π ) ( 0.05 ) 2 − ( 0 ) 2 =0.5π m/s

Thus, the acceleration of the body is 0 m/ s 2 and the velocity of the body is 0.5π m/s .