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 vibrate in two loops leaving the middle point of the wire between the pulleys at rest.

Answer 1

L = 15 m, mass = 12 gm

Mass per unit length

$m=\frac{12}{1.5}g/m$

$=8\times {10}^{-3}kg/m$

$T=9\times g$

$=90N,\lambda =1.5m$

$f_1=\frac{2}{2L}\sqrt{\left(\frac{T}{m}\right)}$

[for second harmonic, two loops are produced]

$f_1=2f_0=\frac{1}{1.5}\sqrt{(\frac{90}{8}\times {10}^{-3})}$

$=\frac{\left(106.06\right)}{\left(1.5\right)}$

$=70.7Hz\approx 70Hz$