Q43E - University Physics with Modern Physics

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.

Step-by-Step Solution

Answer

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

Step 1: Given Data

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

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

Step 1: Formula of fundamental frequency

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

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

Step 2: Calculate the beat frequency

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

$\begin{array}{c}f=\frac{v}{4}\left(\frac{1}{1.14\text{\hspace{0.17em}m}}-\frac{1}{1.16\text{\hspace{0.17em}m}}\right)\\ f=\frac{344\text{\hspace{0.17em}m}/\text{s}}{4}\times \frac{0.02\text{\hspace{0.17em}m}}{1.14\text{\hspace{0.17em}m\hspace{0.17em}}\times 1.16\text{\hspace{0.17em}m}}\\ f=1.3\text{\hspace{0.17em}Hz}\end{array}$

 

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