Question

An object in simple harmonic motion has an amplitude of 0.360 m and an oscillation period of 0.620 s. Determine the maximum speed of the motion.

Answer 1

3.648 m/s

Step-by-step explanation:

Data provided in the question:

Amplitude, A = 0.360 m

oscillation period, T = 0.620 s

The maximum speed of the motion, V = A × ω

where,

ω = Angular speed

also,

ω = (2π) / T

on substituting the respective values, we get

ω = (2π) / 0.620

or

ω = 10.134 rad / s

therefore,

V = 0.360 × 10.134 = 3.648 m/s