In an experiment on standing waves, a string90 cmlong is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 60 Hz. The mass of the string is 0.044 kg . What tension must the string be under (weights are attached to the other end) if it is to oscillate in four loops?

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, $v=\sqrt{\frac{T}{\mu }}.......\left(2\right)$

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:

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

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

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

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