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)

Moderate

A

40 Hz
B

35 Hz
C

30 Hz
D

25 Hz

Solution

For wire if
$\begin{array}{l} M = mass, ρ = density, A = Area of cross section \\ V = volume, l = length, Δ l = change in length \end{array}$
$T h e n m a s s p e r u n i t l e n g t h m = \frac{M}{l} = \frac{A l ρ}{l} = A ρ$
$And Young's modulus of elasticity Y = \frac{T / A}{Δ l / l}$
$\Rightarrow T = \frac{Y Δ l A}{l}$
Hence lowest frequency of vibration
$\begin{array}{l} n = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{Y (\frac{Δ l}{l}) A}{A ρ}} = \frac{1}{2 l} \sqrt{\frac{Y Δ l}{l ρ}} \\ \Rightarrow n = \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{- 4}}{1 \times 9 \times 10^{3}}} = 35 Hz \end{array}$

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