Question

In Fig.16-42, a string, tied to a sinusoidal oscillator at Pand running over a support at Q, is stretched by a block of mass m. Separation L = 1.20 m, linear density, $\mathbf{\mu }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{m}$and the oscillator frequency,. The amplitude of the motion at Pis small enough for that point to be considered a node. A node also exists atQ. (a) What mass mallows the oscillator to set up the fourth harmonicon the string? (b) What standing wave mode, if any, can be set up if m = 1.00 kg?

Short answer

1. Mass of 0.846 kg allows the oscillator to set up the fourth harmonic on the string
2. No standing wave mode can be set up if 1.0 kg

Step-by-step solution

Given data

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

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

The oscillator frequency, f = 120 Hz

Mass of the block = m

Understanding the concept of harmonic oscillation

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, $f=\frac{n}{2L}\times \left(\frac{\tau }{\mu }\right).........\left(1\right)$

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

The tension of the string, $\tau =mg........\left(3\right)$

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

Step 3(a): Calculation of mass that would set up fourth harmonic oscillation

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

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

Hence, the required value of mass is 0.846 kg

Step 4(b): Calculation of standing wave mode if m = 1.00 kg

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

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

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