Question

a particle is moving in simple harmonic motion with an amplitude of 3.2 cm and a maximum velocity of 15.0 cm/s. assume the initial phase is 0. (include the sign of the value in your answer.) (a) at what position is its velocity 4.5 cm/s? cm (b) what is its velocity when its position is 1.6 cm? cm/s

Answer 1

The position when velocity is 4.5cm/s is ±3.05cm and the velocity when the position is 1.6cm/s is 12.99cm.

The maximum velocity of the particle is 15cm/s and the amplitude is 3.2cm.

The velocity of a particle in simple harmonic motion is given by

V = W√(A²-X²)

Where,

V is the velocity,

A is the amplitude,

X is the position,

W is the angular frequency.

So, for maximum velocity,

V = WA

Putting values,

15 = W(3.2)

W = 4.6875 rad/s

(a) For position when speed is 4.5cm/s.

Putting values,

(4.5) = 4.6875√((3.2)²-(X²))

X = ±3.05cm

So, at ±3.05cm, the velocity will be 4.5cm/s.

(b) For velocity at position of 1.6cm,

Putting values,

V = 4.6875√((3.2)+(1.6))

V = 12.99cm/s

So, the velocity is 12.99cm/s at 1.6cm

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