Question

A string (fixed at both ends) of length $1\text{m}$ has the mass per unit length $0.1{\text{g cm}}^{-1}$. What would be the frequency of vibration of this string for third overtone under the tension of $400\text{N}$ ?

• A. $400\text{Hz}$
• B. $300\text{Hz}$
• C. $50\text{Hz}$
• D. $200\text{Hz}$

Answer 1

The correct option is A $400\text{Hz}$

Mass per unit length of string.
$\mu =0.1\text{g/cm}=\frac{0.1\times {10}^{-3}}{{10}^{-2}}=0.01\text{kg/m}$
speed of wave is,
$\Rightarrow v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{400}{0.01}}=200\text{m/s}$
frequency of vibration for string fixed at both ends.
$\Rightarrow f_x=\frac{nv}{2L}$
For 3rd overtone, $n=4\left(4^{th}\text{harmonic}\right)$
$\therefore f_4=\frac{4\times 200}{2\times 1}=400\text{Hz}$