Question 260: Waves

Question

The lengths of the two organ pipes open at both ends are ' L ' and $\left(\mathrm{L}+\mathrm{L}_1\right)$. If they are sounded together, the beat frequency will be ( $\mathrm{v}=$ velocity of sound in air)

Accepted Answer

We are asked:
Two organ pipes, both open at ends, of lengths:
Pipe 1: $L$
Pipe 2: $L + L_1$
and we want the beat frequency.
Step 1: Fundamental frequency of an open pipe
For an open organ pipe of length $L$:
$ f = \frac{v}{2L} $
So:
For pipe 1:
$ f_1 = \frac{v}{2L} $
For pipe 2:
$ f_2 = \frac{v}{2(L+L_1)} $
Step 2: Beat frequency
Difference in frequencies:
$ f_b = |f_1 - f_2| = \left|\frac{v}{2L} - \frac{v}{2(L+L_1)}\right| $
$ f_b = \frac{v}{2}\left|\frac{1}{L} - \frac{1}{L+L_1}\right| $
$ f_b = \frac{v}{2}\cdot \frac{L+L_1 - L}{L(L+L_1)} $
$ f_b = \frac{v}{2}\cdot \frac{L_1}{L(L+L_1)} $
$ f_b = \frac{v L_1}{2L(L+L_1)} $
Final Answer:
$ \boxed{\frac{v L_1}{2L(L+L_1)}} $
That corresponds to
Option D
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Answer

$\frac{2 v L_1}{L\left(L+L_1\right)}$

Answer

$\frac{2 \mathrm{~L}\left(\mathrm{~L}+\mathrm{L}_1\right)}{\mathrm{vL}_1}$

Answer

$\frac{\mathrm{vL}_1}{\mathrm{~L}\left(\mathrm{~L}+\mathrm{L}_1\right)}$

Answer

$\frac{v L_1}{2 L\left(L+L_1\right)}$

Waves

The lengths of the two organ pipes open at both ends are ' L ' and \left(\mathrm{L}+\mathrm{L}_1\right). If they are sounded together, the beat frequency will be ( \mathrm{v}= velocity of sound in air)

Choose the correct answer: