Question

The fundamental frequency on a aluminum string that is fixed at both ends is 468 Hz. The string is then cooled 174°C and as such its length changes. Determine how much the fundamental frequency will change as a result assuming the tension remains constant.

Δf1 =

Answer 1

The fundamental frequency of string that is fixed at both ends

$$f_1=\frac{v}{2L}.$$

Since the tension remains constant, the speed of wave remains constant too.

So

$$f_1'=\frac{v}{2L'}.$$

The linear thermal expansion of string is described by equation

$$L'=L(1+\alpha \Delta T), \; \alpha=25\times 10^{-6}\frac{1}{\degree \rm C}$$

Hence, the change of fundamental frequency due thermal expansion of string

$$\Delta f_1=\frac{v}{2L(1+\alpha \Delta T)}-\frac{v}{2L}=-f_1\alpha \Delta T$$

$$\Delta f_1=-468\times 25\times 10^{-6}\times (-174)=2.0\:\rm Hz.$$