Hi Ray :-)

Hello sir, I had a small doubt, I hope it's not an inconvinience

What's the question?

Suppose, I have a a rod hung by two wires on both ends. Now I add some additional mass on the rod, at some point which is not at the center. It has been told to me that, if I use a tuning fork to excite both the wires, the first one will vibrate at the fundamental frequency and the second one at the first overtone.

I had a minor disagreement

My point was, since the same tuning fork is used, the frequency should be the same in both cases. However, the first wire is vibrating at it's fundamental frequency, and second one is vibrating at the first overtone, so, the tension in the first wire should be more

since f=n/2l sqrt{T l /m}2

In the first wire, f_1=1/2l sqrt{T_1 l/m}, in the second wire it is f_2=2/2l sqrt{T_2 l/m}.

Now I've argued that the same tuning fork excites both wires so f_1 = f_2, and thus T_1>T_2

is that line of reasoning correct ?

This is only going to work with the mass at the correct point on the wire. The frequency is proportional to √T, so you could only get a wavelength L/2 on one string and L on the other if the tension on one string is four times the tension on the other.

yes, I was supposed to find that point on the wire, and I did it using my method

Ah, OK.

yes, but the confusion is regarding which string

as in

is the tension in the first string more, or is it more in the second one

the first string is vibrating at fundamental, and second one is vibrating at first overtone

Suppose the string length is L, then the statement tells us that λ₁ = 2L and λ₂= L.

however, can I truly claim that the overall freqency is equal in both strings, since the same fork excites them.

@RayPalmer Yes, we know f₁ = f₂ = f, where f is the frequency of the tuning fork.

ah, thank you so much, I was trying to get this

OK :-)

So, the tension in the first one must be more

thank you so much