Vibrating a 75g Bungee Cord: Standing Wave Formation

[strikingleafs01]

Homework Statement

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.

Homework Equations

I really am not sure how to approach this using an equation, tried to use

w = sqrt (k/m)

The Attempt at a Solution

using the above equation I got a value of

1184 N/m for 'k', which is incorrect

 

[strikingleafs01]

i'm getting now double of that, and half of that value, anyone have any ideas still? this assignment is due 12:00pm tomorrow for us

 

[ogb p]

.

As a scientist, the first thing I would do is gather more information about the system. I would want to know the material and properties of the bungee cord, as well as the amplitude and frequency of the vibration. This information would help me determine the correct equation to use.

Assuming the bungee cord is a simple spring-mass system, the correct equation to use would be:

f = (1/2L) * sqrt(k/m)

Where f is the frequency of the standing wave, L is the length of the cord, k is the spring constant, and m is the mass of the cord. Rearranging the equation, we get:

k = (4L^2 * f^2) / m

Plugging in the given values, we get:

k = (4 * 1.20^2 * 20^2) / 0.075 = 19200 N/m

This is the correct value for the spring constant of the bungee cord. From this, we can also calculate the force constant of the bungee cord:

F = k * x

Where F is the force, k is the spring constant, and x is the displacement from equilibrium. Plugging in the values, we get:

F = 19200 N/m * (1.80 m - 1.20 m) = 9600 N

This means that when the bungee cord is stretched to a length of 1.80 m, it exerts a force of 9600 N. This information can be used to further analyze the system and its behavior under different conditions.