Question

A string is 0,400m long and has a mass per unit length of 9.00 x 10^-3 kg/m. What must be the tension in the string if it’s second harmonic has the same frequency as the second resonance mode of a 1.75m long pipe open at one end?

Answer 1

Let's first find the wavelength of the wave in the pipe open at one end:

$$\lambda=\dfrac{4}{3}L_{pipe}=\dfrac{4}{3}\cdot1.75\ m=2.33\ m.$$

Then, we can find the frequency from the wave speed equation:

$$f=\dfrac{v}{\lambda}=\dfrac{340\ \dfrac{m}{s}}{2.33\ m}=146\ Hz.$$

Let's find the fundamental frequency of the string:

$$f_1=\dfrac{f_2}{2}=\dfrac{146\ Hz}{2}=73\ Hz.$$

We can find the tension in the string from the formula:

$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}},$$ $$T=4L^2\mu f_1^2,$$ $$T=4\cdot(0.4\ m)^2\cdot9.0\cdot10^{-3}\ \dfrac{kg}{m}\cdot(73\ Hz)^2=30.7\ N.$$