MHT CET · Physics · Waves and Sound
When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with \(0.97 \mathrm{~m}\) length of sonometer wire and the other is in unison with \(0.96 \mathrm{~m}\) length of the same wire. The frequencies of the two tuning forks are
- A \(383 \mathrm{~Hz}, 388 \mathrm{~Hz}\)
- B \(388 \mathrm{~Hz}, 392 \mathrm{~Hz}\)
- C \(475 \mathrm{~Hz}, 480 \mathrm{~Hz}\)
- D \(480 \mathrm{~Hz}, 485 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(D) \(480 \mathrm{~Hz}, 485 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Given number of beats \(=5, l_1=0.97 \mathrm{~m}\), \(l_2=0.96 \mathrm{~m}\)
The frequencies of the given sonometer are as follows:
\(\mathrm{f}_1=\frac{1}{2 \mathrm{l}_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\)
\(\mathrm{f}_2=\frac{1}{2 \mathrm{l}_2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\)
\(\mathrm{f}_2-\mathrm{f}_1=5\)
\(\frac{1}{21_2} \sqrt{\frac{T}{m}}-\frac{1}{21_1} \sqrt{\frac{T}{m}}=5\)
\(\left(\frac{1}{2(0.96)}-\frac{1}{2(0.97)}\right) \sqrt{\frac{T}{m}}=5\)
\(\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=931.2\)
\(\therefore \quad \mathrm{f}_1=480 \mathrm{~Hz}\) and \(\mathrm{f}_2=485 \mathrm{~Hz}\)
The frequencies of the given sonometer are as follows:
\(\mathrm{f}_1=\frac{1}{2 \mathrm{l}_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\)
\(\mathrm{f}_2=\frac{1}{2 \mathrm{l}_2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\)
\(\mathrm{f}_2-\mathrm{f}_1=5\)
\(\frac{1}{21_2} \sqrt{\frac{T}{m}}-\frac{1}{21_1} \sqrt{\frac{T}{m}}=5\)
\(\left(\frac{1}{2(0.96)}-\frac{1}{2(0.97)}\right) \sqrt{\frac{T}{m}}=5\)
\(\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=931.2\)
\(\therefore \quad \mathrm{f}_1=480 \mathrm{~Hz}\) and \(\mathrm{f}_2=485 \mathrm{~Hz}\)
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