# PHYSICS

A 75 g bungee cord has an equilibrium length of 1.20 m. The cord is stretched to a length of
1.80 m, then vibrated at 20 Hz. This produces a standing wave with two antinodes. What is the
spring constant of the bungee cord?

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1. A standing wave with two antinodes has a wavelength for traveling waves equal to the cord length, 1.80 m.

The wave speed V is the product of wavelength and frequency, or 36 m/s

From that wave speed, derive the Tension required to stretch the cord 0.60 m. From that you can get the spring constant.

Note that the lineal density of the stretched cord is
d = 0.075 kg/1.80 m
= 4.167*10^-2 kg/m.

You will need that to relate the wave speed to the tension.

V = sqrt(T/d)

T = d*V^2 = 36^2*4.167*10^-2 = 54 N

Spring constant = 54N/0.60 m = 90 N/m

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posted by drwls

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