Question

What is the fractional change in tension necessary in a sonometer wire of fixed length to produce a note one octave lower than before?
A. $3/4$
B. $2/3$
C. $1/2$
D. $1/4$

Answer 1

Hint: When string is fixed from both ends nodes will be formed at both ends. When a string is only fixed from one end and set free at the other end node is formed at fixed end and antinode is formed at the free end. Since tension and linear mass density are given it is possible to find out the velocity of the standing wave in the string.

Formula used:
$v = \sqrt {\dfrac{T}{\mu }} $
$v = f\lambda $

Complete step by step solution:
When a wave of wavelength lambda is passing through wire with velocity ‘v’ and the tension is T in it, the relation between them is given as $v = f\lambda $.
When string is fixed at both ends initially one loop will be formed and hence wavelength of that loop be ${\lambda _1}$
For one loop distance is $\dfrac{{{\lambda _1}}}{2}$
So initially $\dfrac{{{\lambda _1}}}{2} = l$ , ${f_1} = \dfrac{v}{{{\lambda _1}}} = \dfrac{v}{{2l}}$ … eq1
Then next it forms two loops this time wave length be ${\lambda _2}$
So distance would be ${\lambda _2} = l$, ${f_2} = \dfrac{v}{{{\lambda _2}}} = \dfrac{v}{l} = \dfrac{{2v}}{{2l}}$ … eq2
Then it forms three loops now wavelength is ${\lambda _3}$
So distance would be $\dfrac{{3{\lambda _3}}}{2} = l$, ${f_3} = \dfrac{v}{{{\lambda _3}}} = \dfrac{v}{{\dfrac{{2l}}{3}}} = \dfrac{{3v}}{{2l}}$ … eq3
From equation 1 and 2 and 3 we get
${f_n} = \dfrac{{nv}}{{2l}}$
Velocity of wave would be $\sqrt {\dfrac{T}{\mu }} {\text{ }}$
Where T is tension and $\mu $ is linear mass density
So $f \propto \sqrt T $
One octave higher means frequency gets doubled.
$\eqalign{
  & \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }} \cr
  & \Rightarrow {f_2} = \dfrac{{{f_1}}}{2} \cr
  & \Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = 2 \cr
  & \Rightarrow \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }} = 2 \cr
  & \therefore {T_1} = 4{T_2} \cr} $
As frequency decreased by an octave, tension got decreased by 4 times.
So fractional change would be
$\dfrac{{\Delta T}}{T} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \dfrac{{{T_1} - \left( {\dfrac{{{T_1}}}{4}} \right)}}{{{T_1}}} = \dfrac{3}{4}$

Hence option A would be the answer.

Note:
In case of string fixed at both ends or fixed at one end length of the string is constant but wavelengths are varying and hence frequencies vary. In both ends the fixed case ratio of frequencies will be 1:2:3:4:5….. And so on. So without calculations if we find the first fundamental frequency then we can find the other frequencies.