The length of the wire shown in figure 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.
Solution
Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm
\[\text{ Mass per unit length,} m = \frac{12}{1 . 5} g/m\]
\[ = 8 \times {10}^{- 3} kg/m\]
\[Tension in the wire, T = 9 \times g\]
\[ = 90 \text{ N }\]
Fundamental frequency is given by:
\[f_0 = \frac{1}{2L} \sqrt{\left( \frac{T}{m} \right)}\]
For second harmonic (when two loops are produced):
\[f_1 = 2 f_0 = \frac{1}{1 . 5} \sqrt{\left( \frac{90}{8} \times {10}^{- 3} \right)}\]
\[ = \frac{\left( 106 . 06 \right)}{1 . 5}\]
\[ = 70 . 7 Hz \approx 70 Hz\]
