Frequency of Wave on a Guitar String HW

[slw12]

Homework Statement

When a guitar string is stretched to have a tension T, it produces a frequency f. You change the tension by a very small amount ∆T . Show that the new frequency of the guitar string is

fnew = f ( 1 + (delta T)/2T)

For example, a guitar string has tension T = 10N and produces f = 1000Hz. If you changed the tension by ∆T = 0.01N, what would the new frequency be? Hint: You need calculus for this. Find an expression for f in terms of T. Then take the derivative of f with respect to T.

Homework Equations

fnew = f ( 1 + (delta T)/2T)

(wavelength)(frequency) = √T/μ

The Attempt at a Solution

I have tried taking the derivative of f =(1/wavelength)(√T/μ), but I can't seem to get back to the equation the question acts me to get back to. What am I doing wrong? [/B]

 

[gneill]

Hi slw12,

Welcome to Physics Forums.

Can you show us the details of what you've tried? (PF rules). Hint: you may want to use an expression for T in terms of the original frequency f as a substitution along the way.

 

[haruspex]

I have tried taking the derivative

So what did you get?

The other equation you need is the standard one relating a derivative to the consequences of small changes in the independent variable (Δf, f, Δx, f'.)

 

[slw12]

I have taken the derivative of the 2nd equation above and I got df/dt = (1/(λ*√μ)(1/√T). But I don't know how to get ΔT into the equation

 

[haruspex]

df/dt = (1/(λ*√μ)(1/√T).

Not quite right. What is the derivative of xⁿ wrt x?

how to get ΔT into the equation

Did my hint not help? f(x+Δx)≈ f(x)+ ...?

 

[slw12]

Not quite right. What is the derivative of xⁿ wrt x?

Did my hint not help? f(x+Δx)≈ f(x)+ ...?

Would it be f(x +Δx) = f(x) + Δy?
If that's right, how would I get ΔT/2T as Δy?

 

[haruspex]

Would it be f(x +Δx) = f(x) + Δy?

You are not familiar with Taylor's expansion?
f(x +Δx) = f(x) + Δxf'(x)+ ...
Look it up.