Why fundamental frequency change doesn't match the expected formula

[e(ho0n3]

[SOLVED] Change in Tension & Fundamental Frequency of a String

Problem. Show that if the tension in a streched string is change by a small amount $\Delta F_T$, the frequency of the fundamental is changed by a small amount $\Delta f = 1/2 (\Delta F_T / F_T) f$.

Let T be the intial tension and h the change in tension. The velocity of a transverse wave on the string is $v = \sqrt{T/\mu}$. The initial frequency is

$$f = \frac{v}{\lambda} = \frac{\sqrt{T}}{\lambda \sqrt{\mu}}$$

The new frequency f' is

$$f' = \frac{v'}{\lambda} = \frac{\sqrt{T + h}}{\lambda \sqrt{\mu}}$$

The difference is:

$$f' - f = \frac{1}{\lambda \sqrt{\mu}} \, (\sqrt{T + h} - \sqrt{T})$$

That looks nothing like what I'm trying to show. Now, if I multiply the RHS by $\sqrt{T} / \sqrt{T}$, I get

$$f' - f = \frac{\sqrt{T + h} - \sqrt{T}}{\sqrt{T}} \, f$$

and if I do it again, I get

$$f' - f = \frac{\sqrt{T(T + h)} - T}{T} \, f$$

which is as close as I could get to what needs to be shown.

 

[Doc Al]

binomial expansion

$$\sqrt{T + h} = \sqrt{T}(1 + h/T)^{1/2}$$

Hint: Approximate that expression by taking a binomial expansion to first order in h/T. (Note that h/T << 1)

 

[e(ho0n3]

Hint: Approximate that expression by taking a binomial expansion to first order in h/T. (Note that h/T << 1)

Great hint! I never considered it. The approximation is given below:

$$\sqrt{T(T + h)} = T + 1/2 \, h$$

and so

$$f' - f = \frac{T + 1/2 \, h - T}{T} \, f = \frac{h}{2T} \, f$$

Thanks.