Question

A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.

Answer 1

$m=\left(\frac{4}{80}\right)g/cm$

$=0.005kg/m$

$T=50N$

$L=80cm=0.8m$

$v=\sqrt{\left(\frac{T}{m}\right)}$

$=\sqrt{\left(\frac{50}{0.005}\right)}=100m/s$

Fundamental frequency,

$f=\frac{1}{2L}\sqrt{\left(\frac{T}{m}\right)}$

$=\{\frac{1}{2\times 0.8}\times \sqrt{\left(\frac{50}{0.005}\right)}\}$

$=\frac{100}{(2\times 0.8)}$

$=\frac{100}{16}=62.5Hz$

First harmonic = 62.5 Hz

$f_4=$ frequency of forth harmonic

$=4f_0=f_3=62.5\times 4$

$=250Hz$

$v=f_4{\lambda }_4$

$\Rightarrow \left(\frac{v}{f}\right)=\frac{100}{250}$

$=0.4m=40cm$