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Ahat [919]
2 years ago
12

A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is

the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Physics
1 answer:
andre [41]2 years ago
8 0

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

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Explanation:

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