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: a) amplitude is 45cm (0.45m), b) 3.31 rad/s, c) v = 1.489m/s

Explanation: A)

The amplitude is simply the maximum displacement of an object undergoing simple harmonic motion from it mean position.

From the question, the height from the equilibrium position is 45cm, hence the amplitude is 45cm (0.45m)

B) ω = 2πf

But f = 1/T

Where T is period which is the time taken to complete an oscillation = 1.90s

f = 1/1.9 = 0.526 Hz

But ω = 2πf

ω = 2π (0.526)

ω = 2*3.142*0.526 = 3.31 rad/s

C)

v = ωA

v = 3.31*0.45

v = 1.489 m/s