Question

The fundamental frequency of a stretched string increases by $20$ Hz. Then its tension is increased by $69\%$. Initial fundamental frequency of the string is
A. $25$ Hz
B. $\dfrac{{100}}{3}$ Hz
C. $\dfrac{{200}}{3}$ Hz
D. $50$ Hz

Answer 1

We can find the initial fundamental frequency by using Law of Tension in transverse vibration of a stretched string, which states that the fundamental frequency of transverse vibrations of a stretched string is proportional to the square root of the tension in string.

Formula used:
Law of tension is given by
$n$ $\alpha$ $\sqrt T$
$\dfrac{{{n_1}}}{{\sqrt {{T_1}} }} = \dfrac{{{n_2}}}{{\sqrt {{T_2}} }} =$ constant
Also, $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }}$
Where, $n$ - fundamental frequency of transverse vibration of a stretched string and $T$ - tension in the string.

Complete step by step answer:
Let us assign some terminologies to the given data for better understanding.
${n_1}$ - Initial fundamental frequency of the string
${n_2}$ - Final frequency of the string
${T_1}$ - Initial tension in the string
${T_2}$ - Final tension in the string

We have to calculate the value of ${n_1}$. Getting back to the given conditions in the questions, the fundamental frequency of a stretched string increases by $20$ Hz and tension increases by $69\%$.
Thus,
${n_2} = {n_1} + 20$
$\Rightarrow{T_2} = (1 + 0.69){T_1} = 1.69{T_1}$
Now, using the formula for law of tension stated above, we have
$\dfrac{{{n_1}}}{{\sqrt {{T_1}} }} = \dfrac{{{n_2}}}{{\sqrt {{T_2}} }}$

Substituting the values of ${n_2}$ and ${T_2}$ , we get
$\dfrac{{{n_1}}}{{\sqrt {{T_1}} }} = \dfrac{{{n_1} + 20}}{{\sqrt {1.69{T_1}} }}$
$\Rightarrow \dfrac{{{n_1}}}{{\sqrt {{T_1}} }} = \dfrac{{{n_1} + 20}}{{1.3\sqrt {{T_1}} }} \\\$
$\sqrt {{T_1}}$gets cancelled from both sides and we get
$\Rightarrow 1.3{n_1} = {n_1} + 20$
$\Rightarrow 1.3{n_1} - {n_1} = 20$
$\Rightarrow 0.3{n_1} = 20$
$\therefore {n_1} = \dfrac{{20}}{{0.3}}$
Thus, ${n_1} = \dfrac{{200}}{3}$ Hz
The initial fundamental frequency of the stretched string is $\dfrac{{200}}{3}$ Hz.

Hence, option C is correct.

Note: The law of tension is valid only when the linear density and vibrating length of the string are kept constant. Convert the percentage of increased tension into fractions, otherwise you will get different answers for the same question. You can also solve the question using ${n^2}$ $\alpha$ $T$.