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 50N. The frequency of the fourth harmonic of the fundamental is

• A. 0.25 Hz
• B. 25 Hz
• C. 250 Hz
• D. 2500 Hz

Answer 1

The correct option is C

250 Hz

Given mass $=4g=4\times {10}^{-3}kg$

length = 80 cm = 0.8 m

$\mu =\frac{mass}{length}=\frac{4\times {10}^{-3}}{0.8}=0.5\times {10}^{-2}$

given T = 50 N

Frequency of nth harmonic is given by $f_n=\frac{nv}{2L}$

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{50}{0.5\times {10}^{-2}}}$

$=\sqrt{{10}^4}={10}^{2}m{s}^{-1}$

$f_4=\frac{4\times {10}^2}{2\times 0.8}$

$=0.25\times {10}^3=250Hz$