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A 64.0 kg daredevil exhibitionist hangs on to the end of a long bungee cord attached to the railing of the Brooklyn Bridge and oscillates vertically. (Do NOT try this!) The spring constant of the bungee cord is 113 N/m, and the system's damping constant is 0.113 kg/s.
How long does it take for the oscillation amplitude to decrease to 93.8% of its initial value?
What is the frequency of oscillation?

Respuesta :

Samueloladosu37

Answer: t=72.5secs and f= 0.194Hz

Explanation:

0.938=e^[-bt/(2M)]

ln(0.938) = -(0.113 kg/s)(t) / [2(64.0 kg)]

t = 72.5 s

ω = sqrt[k/m - {b/(2M)}²]

ω = sqrt[(113 N/m)/(64.0 kg) - {(0.113kg/s)/(2(64.0 kg))}²]

ω = 1.33 rad/s

2πf = 1.33 rad/s

f = 0.194 Hz

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