When the tension in string is increased by $3 \mathrm{~kg} \omega \mathrm{t}$, the frequency of the…
When the tension in string is increased by $3 \mathrm{~kg} \omega \mathrm{t}$, the frequency of the fundamental mode increases in the ratio $2: 3$. The initial tension in the string is
$1.6 \mathrm{~kg} \omega \mathrm{t}$
$2.0 \mathrm{~kg} \omega \mathrm{t}$
$2.4 \mathrm{~kg} \omega \mathrm{t}$
$2.8 \mathrm{~kg} \omega \mathrm{t}$
Solution
The formula for frequency is $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}}}$ After increasing the tension, the frequency becomes $\mathrm{f}^{\prime}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}+3}{\mathrm{~m}}}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\sqrt{\frac{\mathrm{T}}{\mathrm{~T}+3}} \\
\therefore & \frac{2}{3}=\sqrt{\frac{\mathrm{T}}{\mathrm{~T}+3}} \Rightarrow \frac{4}{9}=\frac{\mathrm{T}}{\mathrm{~T}+3} \Rightarrow 4 \mathrm{~T}+12=9 \mathrm{~T} \\
\therefore & 5 \mathrm{~T}=12 \\
\therefore & \mathrm{~T}=2.4 \mathrm{~kg} \omega \mathrm{t}
\end{array}$
.