91Ó°ÊÓ

Length of string is,$L=90\mathrm{cm}\mathrm{or}0.90\mathrm{m}$

The frequency of the tuning fork is,$f=60\mathrm{Hz}$

Mass of the string is,$m=0.044\mathrm{kg}$.

We know the formula for the frequency of the standing wave to oscillate in n loops in terms of the velocity of the standing wave (v). We also know the formula for combining these two formulae. Rearranging them, we can get an expression for the tension in the string. Inserting the values in it, we can find its value.

Formula:

The frequency of standing wave for n loops of oscillation,$f=\frac{nv}{2L}.......\left(1\right)$

The velocity of the body, $v=\sqrt{\frac{T}{\mathrm{μ}}}.......\left(2\right)$

Frequency of standing wave to oscillate in n loops

$f=\frac{nv}{2L}$

Using equation (2), the velocity of the string can be given as:

$v=\sqrt{\frac{T}{\left({\displaystyle \frac{m}{L}}\right)}}.........\left(3\right)(âˆ\mu \mathrm{μ}=\frac{m}{L})$

Substituting the value of equation (3) in equation (1), we get the frequency of the oscillation as:

$$\begin{gathered} f=\frac{n\sqrt{{\displaystyle \frac{T}{\left({\displaystyle \frac{m}{L}}\right)}}}}{2L} \\ f=\frac{n\sqrt{{\displaystyle \frac{T}{mL}}}}{2} \\ {f}^2=\frac{n^2\left({\displaystyle \frac{T}{mL}}\right)}{4} \\ T=\frac{4f^{2}mL}{n^2} \\ =\frac{4{\left(60\right)}^2\left(0.044\right)\left(0.90\right)}{4^2} \\ =35.64 \\ â‰\dots 36\mathrm{N} \end{gathered}$$

Therefore, the tension in the string if it is to oscillate in four loops is $36\mathrm{N}$.