Frequency of a Standing Wave on a String

[bcjochim07]

1. The problem statement, all variables and given/known data
The frequency of a standing wave on a string is f when the string's tnesion is T. If the tension is changed by the small amount deltaT, witout changing the length, show tat the frequency changes by an amount deltaf, such that

deltaf/f = .5 * deltaT/T


2. Relevant equations



3. The attempt at a solution

v=sqrt(T/Mu)

f= (1/lambda)*sqrt(T/Mu) When tension is increased, the wavelength will still be the same

f+deltaf=(1/lambda)*sqrt((T+deltaT)/Mu)
so delta f=(1/lambda)*sqrt((T+deltaT)/Mu)-f

deltaf/f = [(1/lambda)*sqrt((T+deltaT)/Mu)-f]/((1/lambda)*sqrt(T/Mu))

deltaf/f = sqrt(T+deltaT)/sqrt(T) -1

But I can't get it simplified any more than this

[bcjochim07]

any thoughts on this one?

[Antenna Guy]

v=sqrt(T/Mu)

f= (1/lambda)*sqrt(T/Mu) When tension is increased, the wavelength will still be the same


If \lambda = \frac{v}{f} and \frac{dv}{dT}=\frac{1}{2\sqrt{T\mu}}, then how is it that the wavelength will be the same when tension is increased?

Regards,

Bill

[bcjochim07]

It seems to me that if both f and v increase by some factor, that wavelength should remain the same, and if there is one wave still on the string when the tension changes very slightly, how could the wavelength change?

[Antenna Guy]

how could the wavelength change?

Because \frac{d\lambda}{dT}\neq 0. Therefore, both \lambda and f are functions of T.

Regards,

Bill