Question

A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f = 680Hz. For what values of L will the tube resonate with the speaker?

Answer 1

L is the length of the tube

f is the frequency

$\lambda$ is the wavelength

$\dfrac{\lambda}{2}=d_{\lambda\lambda}=\dfrac{L}{n}$

$\Rightarrow L= \dfrac{n\lambda}{2}$ for n= 1,2,3, K

But, we know that

$\lambda = \dfrac{v}{f}$

Putting the values of $\lambda,$ we get

$L=n(\dfrac{v}{2f})$ for n= 1,2,3, K

$v=343$ m/s and f= 680 Hz

On substituting, we get

$L =n(\dfrac{343m/s}{2\times 680Hz})=n(0.252)$

for n = 1,2,3,K

Hence, the value of L will be 0.252m , 0.504m, 0.756m ....