MHT CET · Physics · Waves and Sound
A sonometer wire \(49 \mathrm{~cm}\) long is in unison with a tuning fork of frequency ' \(n\) '. If the length of the wire is decreased by \(1 \mathrm{~cm}\) and it is vibrated with the same tuning fork, 6 beats are heard per second. The value of ' \(n\) ' is
- A \(256 Hz\)
- B \(288 Hz\)
- C \(320 Hz\)
- D \(384 Hz\)
Answer & Solution
Correct Answer
(B) \(288 Hz\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2 \\
\therefore & \mathrm{n}_1 \times 49=\mathrm{n}_2 \times 48 \\
\therefore & \mathrm{n}_2=\frac{49}{48} \mathrm{n}_1
\end{array}\)
Given that, beat frequency, \(\left|\mathrm{n}_1-\mathrm{n}_2\right|=6\)
\(\begin{array}{ll}
\therefore & \left|\mathrm{n}_1-\frac{49}{48} \mathrm{n}_1\right|=6 \\
\therefore & \frac{\mathrm{n}_1}{48}=6 \\
\therefore & \mathrm{n}_1=288 \mathrm{~Hz}
\end{array}\)
& \mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2 \\
\therefore & \mathrm{n}_1 \times 49=\mathrm{n}_2 \times 48 \\
\therefore & \mathrm{n}_2=\frac{49}{48} \mathrm{n}_1
\end{array}\)
Given that, beat frequency, \(\left|\mathrm{n}_1-\mathrm{n}_2\right|=6\)
\(\begin{array}{ll}
\therefore & \left|\mathrm{n}_1-\frac{49}{48} \mathrm{n}_1\right|=6 \\
\therefore & \frac{\mathrm{n}_1}{48}=6 \\
\therefore & \mathrm{n}_1=288 \mathrm{~Hz}
\end{array}\)
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