The fundamental frequency of a sonometer wire is $50 \mathrm{~Hz}$ for some length and tension. If the…

The fundamental frequency of a sonometer wire is $50 \mathrm{~Hz}$ for some length and tension. If the length is increased by $25 \%$ by keeping tension same, then frequency change of second harmonic is
  1. decreased by $10 \%$
  2. decreased by $20 \%$
  3. decreased by $5 \%$
  4. decreased by $15 \%$

Solution

$\begin{aligned} & \quad \mathrm{n}_2 \ell_2=\mathrm{n}_1 \ell_1 \\ & \therefore \mathrm{n}_2=\frac{\ell_1}{\ell_2} \cdot \mathrm{n}_1=\frac{\ell_1}{1.25 \mathrm{R}_1} \mathrm{n}_1=0.8 \mathrm{n} \\ & \therefore \mathrm{n}_1-\mathrm{n}_2=\mathrm{n}_1(1-08)=0.2 \mathrm{n}_1 \\ & \\ & \quad \frac{\mathrm{n}_1-\mathrm{n}_2}{\mathrm{n}_1}=0.2\end{aligned}$ $0.2 * 100=20 \%$ ~

Asked in: MHT CET 2020 (19 Oct Shift 1)

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