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

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m)

\[= \left( \frac{4}{80} \right)  g/cm   = 0 . 005  kg/m\]

\[\text{ Wave  speed, }  \nu = \sqrt{\left( \frac{T}{m} \right)}\] 

\[ = \sqrt{\left( \frac{50}{0 . 005} \right)} = 100  m/s\]

\[\text{ Fundamental  frequency  ,}    f_o  = \frac{1}{2L}\sqrt{\left( \frac{T}{m} \right)}\] 

\[       = \frac{1}{2 \times 0 . 8} \times \sqrt{\left( \frac{50}{0 . 005} \right)}\] 

\[       = \frac{100}{2 \times 0 . 8} = 62 . 5  Hz\] 

\[\text { First  harmonic = 62 . 5  Hz }\] 

If   f_4  =frequency  of  the  fourth  harmonic:

\[ \Rightarrow  f_4  = 4 f_0  = 62 . 5 \times 4\] 

\[ \Rightarrow  f_4  = 250  Hz\] 

\[\text{ Wavelength  of  thefourth  harmonic,}    \lambda_4  = \frac{\nu}{f_4} = \frac{100}{250}\] 

\[ \Rightarrow  \lambda_4  = 0 . 4  m = 40  cm\]