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MHT CET · Physics · Oscillations

A violin emits sound waves of frequency ' \(\mathrm{n}_1\) ' under tension T. When tension is increased by \(44 \%\), keeping the length and mass per unit length constant, frequency of sound waves becomes ' \(\mathrm{n}_2\) '. The ratio of frequency ' \(\mathrm{n}_2\) ' to frequency ' \(n_1\) ' is

  1. A \(5: 6\)
  2. B \(6: 7\)
  3. C \(6: 5\)
  4. D \(7: 6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(6: 5\)

Step-by-step Solution

Detailed explanation

Frequency,
\(\mathrm{n}_1=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{~T}_1}{\mu}}\)
Tension is increased by \(44 \%\),
\(\begin{aligned}
& \mathrm{T}_2=\mathrm{T}_1+\frac{44}{100} \mathrm{~T}_1=1.44 \mathrm{~T}_1 \\
\therefore \quad \mathrm{n}_2 & =\frac{1}{2 \mathrm{~L}} \sqrt{\frac{1.44 \mathrm{~T}_1}{\mu}} \\
& \frac{\mathrm{n}_2}{\mathrm{n}_1}=\sqrt{\frac{1.44 \mathrm{~T}_1}{\mathrm{~T}_1}}=\frac{6}{5}
\end{aligned}\)