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## Homework Statement

A uniform 165 N bar is supported horizontally by two identical wires A and B. A small 185 N cube of lead is placed 3/4 of the way from A to B. The wires are each o.75 m long and have a mass of 5.50 g. If both of them are simultaneously plucked at the center, what is the frequency of the beats that they will produce when vibrating in their fundamental?

## Homework Equations

fundamental frequency = (1/2L) * √(F/μ).

where L is the length of the wire, (μ) is the mass per unit length, F is the tension in the wire

f(beat) = f1-f2.

## The Attempt at a Solution

I tried to calculate the frequency of each wire A and B.

I let frequency1 be the frequency of the wire with the block closer to it, so wire B.

frequency1=(1/(2*0.75m))*√(F1)/(0.0055kg/0.75m)

frequency1=7.785*√F1

frequency2=7.785*√F2, as L and μ are the same for both wires.

This is where I got stuck. I tried to calculate the tensions:

F1= (1/2)*(165N) + (3/4)*185 = 221.25N

F2= (1/2)*(165N) + (1/4)*185 = 128.75N

I'm not sure if the tension due to the block would be distributed in the way I did.

When i put these tensions into the equation i get a large number, but i know that beats don't occur at frequency differences greater than 10-15 Hz.

Thank you.