Question

A child bounces in a harness suspended from a door frame by three elastic bands.

(a) If each elastic band stretches 0.200 m while supporting a 6.35-kg child, what is the force constant for each elastic band? (N/m)

(b) What is the time for one complete bounce of this child? (s)

(c) What is the child's maximum velocity if the amplitude of her bounce is 0.200 m? (m/s)

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the child = m = 6.35 kg

Force constant for each elastic band = k

Equivalent force constant of the three elastic bands = k_eq

k_eq = 3k

Amount each elastic band stretches by = X = 0.2 m

mg = k_eqX

mg = 3kX

(6.35)(9.81) = 3k(0.2)

k = 103.82 N/m

Time period for one complete bounce of the child = T

The child will be undergoing a simple harmonic motion therefore the time period of oscillation is given by,

T = 0.897 sec

Angular frequency of oscillation =

= 7 rad/s

Amplitude of the bounce of the child = A = 0.2 m

Maximum velocity of the child = V_max

Maximum velocity in a simple harmonic motion is given by,

V_max = A

V_max = (0.2)(7)

V_max = 1.4 m/s

a) Force constant for each elastic band = 103.82 N

b) Time for one complete bounce of this child = 0.897 sec

c) Child's maximum velocity if the amplitude of her bounce is 0.2 m = 1.4 m/s