Question

A wire having a linear density 0.10 kg/m is kept under a tension 490 N it is observed that it resonates at a frequency of 400 Hz and the next frequency 450 Hz. The length of the wire is

• A. 14 m
• B. 0.7 m
• C. 1.6 m
• D. 0.49 m

Answer 1

Answer: (B)

0.7 m

Fundamental frequency of the oscillations is given as:

$f_0=f_2-f_1$

$f_0=450Hz-400Hz$

$f_0=50Hz$

We know that:

$f_0=\frac{v}{2l}$

And $v=\sqrt{\frac{T}{\mu }}$

So, $f_0=\frac{\sqrt{T/\mu }}{2l}$

Therefore, the length of the wire is given as:

$l=\frac{\sqrt{T/\mu }}{2f_0}$

$l=\frac{\sqrt{490/0.1}}{2\times 50}$

$l=\frac{70}{100}=0.7m$