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

Question

3 years ago

12

the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m

1 answer:

andre [41] · Answer 1 · 2022-10-02T04:05:57+03:00

andre [41]3 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