Question

A string has linear mass density, $\mu =0.1{\text{kg m}}^{-1}$, Length $L=60\text{cm}$ is clamped at $A$ and $B$ and is kept under a tension of $T=160\text{N}$. [The tension providing arrangement has not been shown in the figure]. A small paper rider is placed on the string at point $R$ such that $BR=20\text{cm}$. The string is set into vibrations using a tuning fork of frequency $f$.

• A. The speed of wave set up along string is $40{\text{ms}}^{-1}$
• B. The speed of waves set up along string is $20{\text{ms}}^{-1}$
• C. If $R$ is a node, then the frequency of tuning fork, ${}^{\prime }{f}^{\prime }$ may be $100\text{Hz}$.
• D. If $R$ is a antinode, then the frequency of tuning fork, ${}^{\prime }{f}^{\prime }$ may be $50\text{Hz}$.

Answer 1

Answer: (A), (C)

The correct options are
A The speed of wave set up along string is $40{\text{ms}}^{-1}$
C If $R$ is a node, then the frequency of tuning fork, ${}^{\prime }{f}^{\prime }$ may be $100\text{Hz}$.

Wave speed, $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{160}{0.1}}=40{\text{ms}}^{-1}$
If $R$ is node and segment $BR$ has $n$ loops then segment $AR$ must have $2n$ loops [as $AR=2BR$]
$\therefore n\left[\frac{\lambda }{2}\right]=0.2m\Rightarrow n\left[\frac{v}{f}\right]=0.4$
$\Rightarrow f=n\times 100\text{Hz}$ for $n=1,2,3,...,9$
We get $f=100,200,300,...900\text{Hz}$

If $R$ is antinode, then
$BR=n\frac{\lambda }{2}+\frac{\lambda }{4}$ $[n=0,1,2,...]$
$AR=2BR=2n\frac{\lambda }{2}+\frac{\lambda }{2}$
$\therefore AB=3n\frac{\lambda }{2}+\frac{\lambda }{2}+\frac{\lambda }{4}=[3n+1]\frac{\lambda }{2}+\frac{\lambda }{4}$
But length of string can only be integral multiple of $\frac{\lambda }{2}$
Hence, it is not possible to have antinode at $R$.