Question

The length of the wire shown in figure (15-E8) between the pulley is 1⋅5 m and its mass is 12⋅0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

Answer 1

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

$$\begin{gathered} \mathrm{Mass}\mathrm{per}\mathrm{unit}length,m=\frac{12}{1.5}\mathrm{g}/\mathrm{m} \\ =8\times {10}^{-3}\mathrm{kg}/\mathrm{m} \end{gathered}$$

$$\begin{gathered} \mathrm{Tension}\mathrm{in}\mathrm{the}wire,T=9\times g \\ =90\mathrm{N} \end{gathered}$$


Fundamental frequency is given by:
$f_0=\frac{1}{2L}\sqrt{\left(\frac{T}{m}\right)}$
For second harmonic (when two loops are produced):
$$\begin{gathered} f_1=2f_0=\frac{1}{1.5}\sqrt{\left(\frac{90}{8}\times {10}^{-3}\right)} \\ =\frac{\left(106.06\right)}{1.5} \\ =70.7\mathrm{Hz}\approx 70\mathrm{Hz} \end{gathered}$$