Question

A string with a mass density of $4\times {10}^{-3}\text{kg/m}$ is under a tension of $360\text{N}$ and is fixed at both ends. One of its resonance frequencies is $375\text{Hz}$. The next higher resonance frequency is $450\text{Hz}$. If the mass of the string is represented by $x\times {10}^{-3}\text{kg}$, find $x$

Answer 1

Given that,
Mass density of the string, $\mu =4\times {10}^{-3}\text{kg/m}$
Tension, $T=360\text{N}$
Frequency of first resonance, $f_1=375\text{Hz}$
Next higher frequency, $f_2=450\text{Hz}$
We can write,
$f_1=375=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$ and $f_2=450=\frac{n+1}{2l}\sqrt{\frac{T}{\mu }}$
where $n$ is number of segments vibrating.
$\frac{f_2}{f_1}=\frac{450}{375}=\frac{n+1}{n}\Rightarrow n=5$
So, length of string $l=\frac{n}{2\times f_1}\sqrt{\frac{T}{\mu }}=\frac{5}{2\times 375}\sqrt{\frac{360}{4\times {10}^{-3}}}=2\text{m}$
$\Rightarrow$ Mass of wire $=\left(\mu \right)\left(l\right)=(4\times {10}^{-3}\times 2)=8\times {10}^{-3}\text{kg}$