MHT CET · Physics · Waves and Sound
A uniform string is vibrating with a fundamental frequency ' \(n\) '. If radius and length of string both are doubled keeping tension constant then the new frequency of vibration is
- A \(2 \mathrm{n}\)
- B \(3 \mathrm{n}\)
- C \(\frac{n}{4}\)
- D \(\frac{\mathrm{n}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{n}{4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & l_2=2 l_1, \mathrm{R}_2=2 \mathrm{R}_1, \mathrm{~T}_1=\mathrm{T}_2 \\ & \mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ & \quad \text { Where, } \mathrm{m}=\text { mass per unit length }=\frac{\left(\mathrm{nR}^2 l\right) \rho}{l} \\ & \therefore \quad \mathrm{m} \propto \mathrm{R}^2 \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{l_2} \times \frac{\mathrm{R}_1}{\mathrm{R}_2} \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{2 l_1} \times \frac{\mathrm{R}_1}{2 \mathrm{R}_1} \\ & \therefore \quad \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}\end{aligned}\)
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