Calculating Fundamental Frequency and Overtone Frequency of a Stretched Wire

AI Thread Summary
To calculate the fundamental frequency of a stretched wire, the relevant equation is f = (1/2L)√(T/(m/L)), where L is the length of the vibrating segment. In this case, the effective length to use is 50 cm, as specified by the sonometer stops. The mass of the wire is 0.40 g, and the tension applied is 500 N. For the second overtone, the frequency can be determined using the harmonic series based on the fundamental frequency calculated. Clarification on the use of lengths in the equation is necessary, as the total length and the vibrating length may differ.
i'mcheche
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Homework Statement


a flexible wire is 80 cm long has a mass of 0.40g. It is stretched across stops on a sonometer that are 50 cm apart by a force of 500N. Find the fundamental frequency with which the wire may vibrate. If it vibrates in 2 third overtone, what is it's frequency?

Homework Equations


f=\frac{1}{2L}\sqrt{\frac{T}{\frac{m}{L}}}

The Attempt at a Solution


I don't kno where to start. I substituted all the given data. And I'm not really sure about my answer.

I can't decide what length should be used in the equation. the problem states the total length of the wire and the length of the wire which is streched on the sonometer.
 
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welcome to PF :smile:

it is said that the sonometer stops are 50cm apart.So that should me the length. :smile:
 
There are two L's in the equation. Are they necessarily the same?
 
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