Question:
Two tuning forks when sounded together produce 4 beats per second. One of the forks is in unison with 23 cm length of sonometer wire and other with 24 cm length of the same wire. The frequencies of the two tuning forks are
Two tuning forks when sounded together produce 4 beats per second. One of the forks is in unison with 23 cm length of sonometer wire and other with 24 cm length of the same wire. The frequencies of the two tuning forks are
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Frequency of a sonometer wire varies inversely with the vibrating length.
Updated On: Nov 13, 2026
- \(96\text{ }\mathrm{Hz},92\text{ }\mathrm{Hz}\)
- \(92\text{ }\mathrm{Hz},88\text{ }\mathrm{Hz}\)
- \(72\text{ }\mathrm{Hz},68\text{ }\mathrm{Hz}\)
- \(48\text{ }\mathrm{Hz},44\text{ }\mathrm{Hz}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
For a given wire, \(f\propto\dfrac1L\). The shorter length (23 cm) gives the higher frequency.
Step 2: Ratio
\(\dfrac{f_1}{f_2}=\dfrac{24}{23}\) and \(f_1-f_2=4\ \text{Hz}\).
Step 3: Solve
\(f_2\left(\dfrac{24}{23}-1\right)=4\Rightarrow f_2=92\ \text{Hz}\).
Step 4: Other fork
\(f_1=92+4=96\ \text{Hz}\).
Final Answer:
The frequencies are 96 Hz and 92 Hz. \[ \boxed{\text{(A) }96\text{ Hz},92\text{ Hz}} \]
For a given wire, \(f\propto\dfrac1L\). The shorter length (23 cm) gives the higher frequency.
Step 2: Ratio
\(\dfrac{f_1}{f_2}=\dfrac{24}{23}\) and \(f_1-f_2=4\ \text{Hz}\).
Step 3: Solve
\(f_2\left(\dfrac{24}{23}-1\right)=4\Rightarrow f_2=92\ \text{Hz}\).
Step 4: Other fork
\(f_1=92+4=96\ \text{Hz}\).
Final Answer:
The frequencies are 96 Hz and 92 Hz. \[ \boxed{\text{(A) }96\text{ Hz},92\text{ Hz}} \]
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