Question:
The lengths of the two organ pipes open at both ends are ' $L$ ' and $(L + L_1)$. If they are sounded together, the beat frequency will be ($v = $ velocity of sound in air)}
The lengths of the two organ pipes open at both ends are ' $L$ ' and $(L + L_1)$. If they are sounded together, the beat frequency will be ($v = $ velocity of sound in air)}
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Beat frequency is the difference between two nearby frequencies: $b = |n_1 - n_2|$.
Updated On: Apr 26, 2026
- $\frac{2vL_1}{L(L+L_1)}$
- $\frac{2 L(L+L_1)}{vL_1}$
- $\frac{vL_1}{L(L+L_1)}$
- $\frac{vL_1}{2L(L+L_1)}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The fundamental frequency of an open organ pipe is $n = \frac{v}{2L}$.
Step 2: Analysis
Frequency of first pipe ($n_1$) $= \frac{v}{2L}$.
Frequency of second pipe ($n_2$) $= \frac{v}{2(L + L_1)}$.
Step 3: Calculation
Beat frequency ($b$) $= n_1 - n_2 = \frac{v}{2L} - \frac{v}{2(L + L_1)}$
$b = \frac{v}{2} \left[ \frac{(L + L_1) - L}{L(L + L_1)} \right] = \frac{v L_1}{2L(L + L_1)}$.
Step 4: Conclusion
Hence, the correct expression is $\frac{vL_1}{2L(L+L_1)}$.
Final Answer: (D)
The fundamental frequency of an open organ pipe is $n = \frac{v}{2L}$.
Step 2: Analysis
Frequency of first pipe ($n_1$) $= \frac{v}{2L}$.
Frequency of second pipe ($n_2$) $= \frac{v}{2(L + L_1)}$.
Step 3: Calculation
Beat frequency ($b$) $= n_1 - n_2 = \frac{v}{2L} - \frac{v}{2(L + L_1)}$
$b = \frac{v}{2} \left[ \frac{(L + L_1) - L}{L(L + L_1)} \right] = \frac{v L_1}{2L(L + L_1)}$.
Step 4: Conclusion
Hence, the correct expression is $\frac{vL_1}{2L(L+L_1)}$.
Final Answer: (D)
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