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 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 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:

(a) 45 cm (b) 3.305 rad/sec (c) 1.4875 m/sec

Explanation:

It is given that the motion is simple harmonic and it takes 1.90 sec to complete a cycle

The frequency is defined as the reciprocal of time taken to complete one cycle so $f=\frac{1}{1.90}=0.5263\ Hz$

(a) As it is given that the height of each bounce above the equilibrium positon is 45 cm so amplitude will be 45 cm

(b) The angular frequency ω is given by $\omega =2\pi f=2\times 3.14\times 0.5263=3.3051$  rad/sec

(c) The maximum speed $V_{max}$ is given by $V_{max}=\omega A$  where ω is angular frequency of the motion and A is the amplitude So $V_{max}=3.305\times 45\times 10^{-2}=1.4875\ m/sec$