91Ó°ÊÓ

The separation between oscillation and the mass, L = 1.2 m

The linear density of the string,$\mathrm{μ}=1.6\mathrm{g}/\mathrm{m}\mathrm{or}1.6×{10}^{-3}\mathrm{kg}/\mathrm{m}$

The oscillator frequency, f = 120 Hz

Mass of the block = m

We know the equation for the frequency of a standing wave and the velocity of the wave. Using both of these equations, we can find the required mass which allows the oscillator to set up the fourth harmonic on the string.

Using the same formulae we used earlier we can find the wave mode which can be set up when m = 1.00 kg

Formulae:

The frequency of nth harmonic oscillation, $f=\frac{n}{2L}×\left(\frac{\mathrm{τ}}{\mathrm{μ}}\right).........\left(1\right)$

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

The tension of the string, $\mathrm{Ï„}=mg........\left(3\right)$

n = 4.......(4) for fourth harmonic

Rearranging equation (1), we get the given form as:

$$\begin{gathered} \frac{2Lf}{n}=\left(\sqrt{\frac{\mathrm{Ï„}}{\mathrm{μ}}}\right) \\ \frac{2Lf}{n}=\left(\sqrt{\frac{mg}{\mathrm{μ}}}\right) \\ {\left(\frac{2Lf}{n}\right)}^2=\frac{mg}{\mathrm{μ}}{\left(\mathrm{we}\mathrm{take}\mathrm{squares}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{m}=\left(\frac{1.6×{10}^{-3}}{9.8}\right)\left(\frac{2×1.2×120}{4}\right)\right)}^2 \\ m=\frac{\mathrm{μ}}{mg}×{\left(\frac{2Lf}{n}\right)}^2 \\ =\left(\frac{1.6×{10}^{-3}}{9.8}\right)\left(\frac{2×1.2×120}{4}\right)\left(âˆ\mu \mathrm{substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right) \\ =0.846\mathrm{kg} \end{gathered}$$

Hence, the required value of mass is 0.846 kg

Again rearranging equation (1), we get the form as given:

$$\begin{gathered} n=2Lf×\left(\sqrt{\frac{\mathrm{μ}}{\mathrm{τ}}}\right) \\ =2Lf×\left(\sqrt{\frac{\mathrm{μ}}{mg}}\right) \\ =2×1.2×120×\left(\sqrt{\frac{1.6×{10}^{-3}}{1.0×9.8}}\right) \\ =3.68 \end{gathered}$$

As the calculated answer is not an integer, hence, no standing wave can be set up on the given wave.