Question:
When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with $0.70 \text{ m}$ length of sonometer wire and another fork is in unison with $0.69 \text{ m}$ length of the same sonometer wire. The frequencies of the two tuning forks are}
When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with $0.70 \text{ m}$ length of sonometer wire and another fork is in unison with $0.69 \text{ m}$ length of the same sonometer wire. The frequencies of the two tuning forks are}
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Shorter wire length always corresponds to a higher frequency.
Updated On: May 12, 2026
- 320 Hz, 326 Hz
- 414 Hz, 420 Hz
- 420 Hz, 426 Hz
- 480 Hz, 486 Hz
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The Correct Option is B
Solution and Explanation
Step 1: Concept
For a sonometer wire, frequency $n \propto 1/l$. Thus $n_1 l_1 = n_2 l_2$.
Step 2: Meaning
Let the frequencies be $n_1$ and $n_2$. Since $l_2 < l_1$, then $n_2 > n_1$. $n_2 - n_1 = 6$.
Step 3: Analysis
$n_1(0.70) = (n_1 + 6)(0.69) \implies 0.70n_1 = 0.69n_1 + 4.14$.
$0.01n_1 = 4.14 \implies n_1 = 414 \text{ Hz}$.
Step 4: Conclusion
$n_2 = 414 + 6 = 420 \text{ Hz}$. Final Answer: (B)
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