MHT CET · Physics · Waves and Sound
The length and diameter of a metal wire used in sonometer is doubled. The fundamental frequency will change from ' \(n\) ' to
- A \(\frac{\mathrm{n}}{4}\)
- B n
- C 2n
- D \(\frac{\mathrm{n}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{n}}{4}\)
Step-by-step Solution
Detailed explanation
The fundamental frequency is given by
\(\mathrm{n}=\frac{1}{2 \mathrm{Cr}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\)
\(\begin{aligned} & \therefore \mathrm{n} \propto \frac{1}{\ell_{\mathrm{r}}} \\ & \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\ell_1 \mathrm{r}_1}{\ell_2 \mathrm{r}_2}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\end{aligned}\)
\(\therefore \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}\)
\(\mathrm{n}=\frac{1}{2 \mathrm{Cr}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\)
\(\begin{aligned} & \therefore \mathrm{n} \propto \frac{1}{\ell_{\mathrm{r}}} \\ & \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\ell_1 \mathrm{r}_1}{\ell_2 \mathrm{r}_2}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\end{aligned}\)
\(\therefore \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}\)
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