A wire having linear density of 0.05 gcm-1 is stretched between t… - Vidyadip

Question

A wire having linear density of 0.05 gcm^-1 is stretched between two rigid supports with a tension of 4.5 × 10⁷ dynes. It is observed that me wire resonates at a frequency of 420Hz. The next higher frequency at which the wire resonates is 490Hz. Find the length of the wire.

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Answer

Let 420Hz be the pth harmonic, then 490Hz is the (p + 1)th harmonic.
Therefore,
$420=\frac{\text{p}}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}\ \dots(\text{i})$
$\therefore490=\frac{\text{p}+1}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}\ \dots(\text{ii})$
Dividing (ii) by (i), we get
$\frac{490}{420}=\frac{\text{p}+1}{\text{p}}$
$\Rightarrow \text{p}=6.$
Substituting this value of P in eqn. (i), we get
$420=\frac{6}{2\text{L}}\times \sqrt{\frac{4.5\times10^7}{0.05}}$
which give $\text{L}=214.3\text{cm}$

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