Question

A wire having a linear mass density ${10}^{-3}$ kg/m is stretched between two rigid supports with a tension of 90 N. The wire resonates at a frequency of 350 Hz. The next higher frequency at which the same wire resonates is 420 Hz. The length of the wire is

• A. 2.1 m
• B. 4.2 m
• C. 8.4 m
  
• D. 6.3 m
  

Answer 1

The correct option is A 2.1 m

Suppose the wire vibrates at 350 Hz in its nth harmonic and at 420 Hz in its (n + 1)th harmonic
$350s^{-1}=\frac{n}{2L}\sqrt{\frac{F}{\mu }}$ ...............(i)
and $420s^{-1}=\frac{(n+1)}{2L}\sqrt{\frac{F}{\mu }}$ ...................(ii)
This gives $\frac{420}{350}=\frac{n+1}{n}$ or n = 5
Putting the value in (i)
$350=\frac{5}{2l}\sqrt{\frac{90}{{10}^{-3}}}\Rightarrow 350=\frac{5}{2l}\times 300\Rightarrow l=\frac{1500}{700}=\frac{15}{7}m=2.1m$