An object is moving with simple harmonic motion. When the object is displaced 5.00 cm from its equilibrium point, its velocity is 15.0 cm/s. Calculate the amplitude of motion if its period of motion is 3.0 s.
A) 13.6 cm
B) 7.1 cm
C) 8.7 cm
D) 5.0 cm
E) 6.55 cm

Question

Quizplus · Accepted Answer

Given, 
Amplitude = ?
Displacement = 5.00 cm
Velocity = 15.0 cm/s
Period = 3.0 s

We know that for an object in simple harmonic motion, the velocity and displacement are related as follows:

v = Aω cos(ωt)

where v is the velocity, A is the amplitude, ω is the angular frequency (ω = 2π/T where T is the period), and t is the time.

At the maximum displacement, the velocity is zero. Therefore,

Aω(cosπ/2) = 0

A = 0 (not possible, since the object is already displaced)

At a displacement of 5 cm, the velocity is 15 cm/s. Therefore,

Aω(cos(arccos(15/ωA))) = 5

Simplifying this expression and using the fact that ω = 2π/T, we get:

A = (5/T)(2/15π)

Substituting T = 3.0 s, we get:

A ≈ 8.7 cm

Therefore, the amplitude of motion is 8.7 cm. The best choice is C.

An Object Is Moving with Simple Harmonic Motion