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.
Figure

Answer 1

Frequency of Vibration is 70 Hz

Explanation:

Given:

Length of the wire between two pulleys (L) = 1.5 m

Mass of the wire = 12 gm

Mass per unit length, m = $\frac {12}{1.5} gm^-^1$

= $8 \times 10 - 3 kgm^-^1$

Mass per unit length, m = $121.5\ gm^-^1 = 8 \times 10^-^3\ kgm^-^1$

Tension in the wire, T = $9 \times g$

= 90 N  

Fundamental frequency is given by:

$f_0 = \frac {1}{2L} \sqrt(\frac {T}{m})$

For second harmonic (when two loops are produced):

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

= $\frac {106.06}{1.5}$

= 70.7 Hz ≈ 70 Hz