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Q.

A wire of density $9 g m / c m^{3}$ and length 1m is stretched to an extension of 0.05 cm. The lowest frequency of transverse vibrations in the wire is (Young’s modulus, $Y = 9 \times 10^{10} \frac{N}{m^{2}}$ )

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a

35 Hz

b

90 Hz

c

45 Hz

d

75 Hz

answer is A.

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Detailed Solution

Lowest frequency = $\frac{V}{2 l}$ $\Rightarrow \frac{V}{2 l} = \frac{\sqrt{T / ρ A}}{2 l}$

$∴ f = \frac{1}{2 l} \sqrt{\frac{Y A Δ l}{ρ l A}}$

$f = \frac{1}{2 l} \sqrt{\frac{Y Δ l}{l} \times \frac{1}{ρ}}$

$f = \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 0.05 \times 10^{- 2}}{1 \times 9000}}$

$∴ f = 35 H z$

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