MHT CET · Physics · Waves and Sound
The fundamental frequency of a wire stretched by \(2 \mathrm{kgwt}\) is \(100 \mathrm{~Hz}\). The weight required to produce its octave is
- A \(12 \mathrm{kgwt}\)
- B 8 kgwt
- C 4 kgwt
- D \(16 \mathrm{kgwt}\)
Answer & Solution
Correct Answer
(B) 8 kgwt
Step-by-step Solution
Detailed explanation
Octave means the new frequency is \(2 n\)
The velocity of sound is proportional to the square-root of the tension in the wire.
The frequency is directly proportional to the velocity of sound in the wire:
\(\begin{aligned} & \therefore n \propto \sqrt{T} \\ & \Rightarrow \frac{n_2}{n_1}=\sqrt{\frac{T_2}{T_1}} \\ & \therefore T_2=T_1\left[\frac{n_2^2}{n_1^2}\right]=4 T_1=8 \mathrm{kgwt}\end{aligned}\)
The velocity of sound is proportional to the square-root of the tension in the wire.
The frequency is directly proportional to the velocity of sound in the wire:
\(\begin{aligned} & \therefore n \propto \sqrt{T} \\ & \Rightarrow \frac{n_2}{n_1}=\sqrt{\frac{T_2}{T_1}} \\ & \therefore T_2=T_1\left[\frac{n_2^2}{n_1^2}\right]=4 T_1=8 \mathrm{kgwt}\end{aligned}\)
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