A baby bounces up and down in her crib. Her mass is 13.0 kg, and the crib mattress can be modeled as a light spring with force constant 788 N/m. (a) The baby soon learns to bounce with maximum amplitude and minimum effort by bending her knees at what frequency? Hz (b) If she were to use the mattress as a trampoline—losing contact with it for part of each cycle—what minimum amplitude of oscillation does she require? cm

(a) The frequency of the oscillation is 1.24 Hz.

(b) The minimum amplitude of oscillation is 16.2 cm.

The given parameters;

The period of the oscillation is calculated as follows;

The frequency of the oscillation is calculated as follows;

The minimum amplitude of oscillation is calculated as follows;

x = 16.2 cm

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a) 0.804second

b) 0.162m

Explanation:

In simple harmonic motion;

Period T = 2π√m/k

Where m is the mass of the object

K is the force constant

Given:

M = 13.0kg

K = 788N/m

T = 2π√13/788

T = 2π√0.0165

T = 2π×0.128

T = 0.804second

Frequency is the reciprocal of the period. F = 1/T

F = 1/0.804

F = 1.244Hertz

b) To get the amplitude x, we will use the relationship F = kx where F is the force exerted by the baby of mass 13kg

x = F/k

Since F = mg

x = mg/k

Assume g = 9.81m/s²

x = (13×9.81)/788

x = 127.53/788

x = 0.162m

The minimum amplitude that she requires is 0.162m

f=ma newton's second law 7x8=56n