Question

A wire having a linear mass density $5.0 \times 10^{-3} kg / m$ is stretched between two rigid supports with a tension of $450\, N.$ The wire resonates at a frequency of $420\, Hz$. The next higher frequency at which the same wire resonates is $480\, Hz$. Find the length of the wire.

• A. 5 m
• B. 2.5 m
• C. 1.5 m
• D. 25 m

Answer 1

Answer: (B)

Suppose the wire vibrates at $420\, Hz$ in its nth harmonic and at $490\, Hz$ in its $( n +1)$ th harmonic.
$420=\frac{ n }{2 L } \sqrt{\frac{T}{\mu}}$ ...(i)
and $490=\frac{(n+1)}{2 L} \sqrt{\frac{T}{\mu}}$ ...(ii)
Subtracting equation (i) from (ii)
$\frac{1}{2 L} \sqrt{\frac{T}{\mu}}=60$
$\Rightarrow L=\frac{1}{2 \times 60} \sqrt{\frac{T}{\mu}}$
$\therefore L =\frac{1}{2 \times 60} \sqrt{\frac{450}{5 \times 10^{-3}}}$
$=2.5\, m$