Question

A piano wire weighing $6g$and having a length of $90cm$ emits a fundamental frequency corresponding to the “Middle C” $(\nu =261.63Hz)$. Find the tension in the wire.

Answer 1

To find the tension in the wire:

Step 1. Given:
Length of the piano wire,$L=90cm=0.90m$
Mass of the wire $M=6g=0.006kg$
Fundamental frequency $(f_0)=261.63Hz$

Let the tension in the wire be $T,$

Step 2. Linear mass density,

$m=\frac{M}{L}=\left(\frac{6}{90}\right)\raisebox{1ex}{$gm$}\!\left/ \!\raisebox{-1ex}{$cm$}\right.$
$$\begin{gathered} \Rightarrow m=\left(\frac{6\times {10}^{-3}}{90\times {10}^{-2}}\right)\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m$}\right. \\ \Rightarrow m=\left(\frac{6}{900}\right)\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m$}\right. \end{gathered}$$

Step 3. Fundamental frequency,

We know that fundamental frequency,$f_0=\left(\frac{1}{2L}\right)\sqrt{\left(\frac{T}{m}\right)}$

$\Rightarrow 261.63=\left(\frac{1}{2\times 0.90}\right)\sqrt{\frac{T\times 900}{6}}$

$\Rightarrow \left(261.63\times 0.18\right)=\sqrt{150T}$

$$\begin{gathered} \Rightarrow 150T={\left(261.63\times 0.18\right)}^2 \\ \Rightarrow T=\frac{{\left(261.63\times 0.18\right)}^2}{150} \end{gathered}$$

$\Rightarrow T=1480N(approx.)$

Hence, the tension in the piano wire is $1480N$ .