A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93…
A piano wire with a diameter of $0.90 \mathrm{~mm}$ is replaced by another wire of diameter $0.93 \mathrm{~mm}$ of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is
$+3 \%$
$-3 \%$
$+3.2 \%$
$-3.2 \%$
Solution
Given, diameter, $d_1=0.9 \mathrm{~mm}=0.9 \times 10^{-3} \mathrm{~m}$
$
\begin{aligned}
& =9 \times 10^{-4} \mathrm{~m} \\
d_2=0.93 \mathrm{~mm} & =9.3 \times 10^{-4} \mathrm{~m}
\end{aligned}
$
Frequency of fundamental tone is given as
$
f=\frac{1}{2 l} \sqrt{\frac{T}{m}}
$
where, $m=$ mass per unit length.
$
=\frac{M}{l}=\frac{V \cdot \rho}{l}=\frac{A \cdot l}{l} \cdot \rho=A \rho
$
$
\Rightarrow \quad m=\frac{\pi d^2 \rho}{4}
$
From Eqs. (i) and (ii), we get
$
\begin{aligned}
& f=\frac{1}{2 l} \sqrt{\frac{T}{\pi d^2 \rho / 4}}=\frac{1}{2 l} \sqrt{\frac{4 T}{\pi d^2 \rho}}=\frac{1}{l} \sqrt{\frac{T}{\pi d^2 \rho}} \\
\Rightarrow & f=\frac{1}{l d} \sqrt{\frac{T}{\pi \rho}} \Rightarrow f \propto \frac{1}{d} \\
\therefore & \frac{f_1}{f_2}=\frac{d_2}{d_1}=\frac{9.3 \times 10^{-4}}{9 \times 10^{-4}}=\frac{9.3}{9} \Rightarrow \frac{f_2}{f_1}=\frac{9}{9.3}
\end{aligned}
$
$\therefore$ Percentage change in frequency
$
\begin{aligned}
& =\frac{f_2-f_1}{f_1} \times 100=\left(\frac{f_2}{f_1}-1\right) \times 100 \\
& =\left(\frac{9}{9.3}-1\right) \times 100=-3.2 \%
\end{aligned}
$