Question

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

• A. $80cm$
• B. $60cm$
• C. $40cm$
• D. $20cm$

Answer 1

Answer: (C)

$40cm$

$m=$ mass per unit length

$=\frac{4\times {10}^{-3}}{80\times {10}^{-2}}=0.005Kg/m$

given $T=50N$

$L=80cm=0.8m$

$\therefore v=\sqrt{\frac{T}{m}}=\sqrt{\frac{50}{0.005}}=100m/sec$

Fundamental frequency

$f_0=\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2\times 0.8}\sqrt{\frac{50}{0.005}}=625Hz$

$\therefore f_4$ Frequency of fourth harmonic

$4f_0=4\times 62.5=250Hz$

As we know

$v_4=f_4{\lambda }_4\therefore {\lambda }_4=\frac{v_4}{f_4}=\frac{100}{250}=0.4m=40cm$