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?
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.
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.