Question

Two organ pipes, open at one end but closed at the other, are each $\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{m}$ long. One is now lengthened by $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{cm}$ . Find the beat frequency that they produce when playing together in their fundamentals.

Short answer

The beat frequency came out to be $1.3\text{\hspace{0.17em}Hz}$.

Step-by-step solution

Given Data

The length of the pipe is-$1.14\text{\hspace{0.17em}m}$

Increase in length- $2.00\text{\hspace{0.17em}cm}$

Formula of fundamental frequency

The formula of fundamental frequency is $\mathbf{f}\mathbf{=}\frac{v}{\lambda }$.

As pipes are closed from the other end, $\mathbf{\lambda }\mathbf{=}\mathbf{4}\mathbf{L}$ .

Calculate the beat frequency

Beat frequency can be expressed as$f=\frac{v}{4}\left(\frac{1}{L_1}-\frac{1}{L_2}\right)$ .

So, the beat frequency is$1.3\text{\hspace{0.17em}Hz}$ .