A 30.0cm long wire having a mass of 10.0g is fixed at the two end… - Vidyadip

Question

A 30.0cm long wire having a mass of 10.0g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = 340m/s.

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Answer

Given, $\text{m}=10\text{g}=10\times10^{-3}\text{kg},\ \text{l}=30\text{cm}=0.3\text{m}$ Let the tension in the string will be = T$\mu=\frac{\text{Mass}}{\text{Unit length}}=33\times10^{-3}\text{kg}$
The fundamental frequency $\Rightarrow\text{n}_0=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}\ \dots(1)$ The fundamental frequency of closed pipe$\Rightarrow\text{n}_0=\Big(\frac{\text{v}}{4\text{l}}\Big)=\frac{340}{4\times50\times10^{-2}}=170\text{Hz}\ \dots(2)$
According equations (1) × (2) we get$170=\frac{1}{2\times30\times10^{-2}}\times\sqrt{\frac{\text{T}}{33\times10^{-3}}}$
$\Rightarrow\text{T}=347\ \text{Newton}$
Hence, then tansion in the wire is 347 N.

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