Question

A stretched string is $1 \,m$ long. Its mass per unit length is $0.5 \,g/m$. It is stretched with a force of $20 \,N$. It is plucked at a distance of $25 \,cm$ from one end. The frequency of note emitted by it will be:

• A. 400 m
• B. 300 Hz
• C. 200 Hz
• D. 100 Hz

Answer 1

Answer: (C)

As stretched string of $1\, m$ long is plucked at a distance of $25\, cm$ from end, so there will be $2$ loops.
Frequency, $n=\frac{p}{2 l} \sqrt{\frac{T}{m}}$
Given, $p=2, l=1 \,m, T=20 \,N$,
$ m=0.5 \times 10^{-3} kg / m$
$ \therefore n=\frac{2}{2 \times 1} \sqrt{\frac{20}{0.5 \times 10^{-3}}}$
$ \Rightarrow n=200\, Hz$