A tube 1.20mlong is closed at one end. A stretched wire is placed near the open end. The wire is 0.330mlong and has a mass of 9.60g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column’s fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

1. Length of the tube is, L = 1.20m.
2. Length of the wire is, l = 0.33m.
3. Mass of the wire is, m = 9.60g.

The frequency of the resonance with n = 1is called the fundamental frequency. It is given that the tube is closed at one end, so, we can use the formula for resonant frequency.

Formula:

The resonant frequency of the body in SHM,

$\mathit{f}\mathbf{=}\mathit{n}\frac{\mathbf{v}}{\mathbf{4}\mathbf{L}}$ …(1)

The velocity of the resonant frequency in SHM,

$\mathit{v}\mathbf{=}{\left(\frac{T}{\mu }\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}$ …(2)

Where, v is the velocity of sound, T is the tension in the wire, and d is the linear density

By resonance, the wire sets the air column in the tube at fundamental frequency. So, the fundamental frequency can be found using equation (1) for n = 1 as: where v = 343m/s is the velocity of sound.

$\begin{array}{rcl}f& =& \frac{343m/s}{4\times 1.20m}\\ & =& \frac{343m/s}{4.8m}\\ & =& 71.458Hz\\ & & \square 71.5Hz\end{array}$

Because of the resonance, the tube andthe wire have the same fundamental frequency 71.5Hz.

Hence, the value of the fundamental frequency is 71.5Hz

Combining equations (1) and (2), we get the tension of the string as:

$\begin{array}{rcl}2lf& =& {\left(\frac{T}{\mu }\right)}^{1/2}\\ 4l^2{f}^2& =& \frac{T}{\mu }\\ \because \mu & =& \frac{m}{l}\end{array}$

$$\begin{gathered} T=4l^2{f}^2\times \mu \\ T=4l^2{f}^2\times \frac{m}{l} \\ T=4ml{f}^2 \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} T=4\times \left(9.60\times {10}^{-3}kg\right)\times 0.33m\times {\left(71.5Hz\right)}^2 \\ T=64.8N \end{gathered}$$

Hence, the value of the tension of the string is, 64.8N.