A wire of density 9 $\times$ 10–3 kg cm–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 $\times$ 10–4. The lowest frequency of the transverse vibrations in the wire is : (Young’s modulus of wire Y = 9 $\times$ 1010 Nm–2), (to the nearest integer), _________
Answer (integer)
35
Solution
$\rho$<sub>wire</sub> = 9 $\times$ 10<sup>–3</sup> kg cm<sup>–3</sup>
<br><br>= ${{9 \times {{10}^{ - 3}}} \over {{{10}^{ - 6}}}}$ kg/m<sup>3</sup>
= 9000 kg/m<sup>2</sup>
<br><br>f = ${1 \over {2l}}\sqrt {{T \over \mu }} =$${1 \over {2l}}\sqrt {{T \over {{\rho _{wire}}A}}}$
<br><br>= ${1 \over {2l}}\sqrt {{{Y\Delta l} \over {{\rho _{wire}} A}}}$
<br><br>$$ = {1 \over {2 \times 1}}\sqrt {{{9 \times {{10}^{10}} \times 4.9 \times {{10}^{ - 4}}} \over {9000 \times 1}}} $$
<br><br>= 35 Hz
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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