First slide

Stationary waves

Question

A wire of density 9 $\times$ 10³ kg /m³ is stretched between two clamps 1 m apart and is subjected to an extension of 4.9 $\times$ 10^-4 m. The lowest frequency of transverse vibration in the wire is (Y = 9 $\times$ 10¹⁰ N / m²)

Difficult

A

40 Hz
B

35 Hz
C

30 Hz
D

25 Hz

Solution

For wire if
M = mass, $ρ$ = density, A = Area of cross section
V = volume, l = length, $Δl$ = change in length
Then mass per unit length $m = \frac{M}{l} = \frac{Alρ}{l} = Aρ$
And Young’s modules of elasticity $y = \frac{T / A}{Δl / l}$
$\Rightarrow T = \frac{YΔlA}{l}$ . Hence lowest frequency of vibration
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{y (\frac{Δl}{l}) A}{Aρ}} = \frac{1}{2 l} \sqrt{\frac{yΔl}{lρ}}$
$\Rightarrow n = \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 4 .9 \times 10^{- 4}}{1 \times 9 \times 10^{3}}} = 35 Hz$

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