Question

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 2.65 s to complete one cycle. The height of each bounce above the equilibrium position is 36.0 cm. Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person

Answer 1

Answer with Explanation:

We are given that

Time period,T=2.65 s

a.Distance between the equilibrium position and the point at  maximum height is called amplitude.

Therefore, A=36 cm=$\frac{36}{100}=0.36 m$

 1 m=100 cm

b.Angular frequency,$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.65}=2.37 rad/s$

c.Maximum speed,$v_{max}=A\omega$

Using the formula

Maximum speed,$v_{max}=0.36\times 2.37=0.85 m/s$