Question

A steel wire of length 1m, mass 0.1 kg and uniform cross sectional area 10^-6m² is rigidly fixed at both ends. The temperature of the wire is lowered by 20°C. If transverse waves are set up by plucking the wire in the middle, what is the frequency of the fundamental mode of vibration?

• A. 25Hz
• B. 28HZ
• C. 11Hz
• D. 15 Hz(Young’s modulus of steel = 2.010¹¹ N/m², coefficient of linear expansion of steel = 1.21 10^-5 per °C)

Answer 1

Answer: (C)

11Hz

Answer 2

The frequency of the fundamental mode of vibration is given by the formula

n = $\frac{1}{2\mathcal{l}}\sqrt{\frac{T}{m}}$, where T is the tension.

The force of tension is provided by the contraction due to the decrease in temperature.

Contraction = αLt = 1.21 10^-6L20

Now, Y = $\frac{T \times 1}{10^{- 6} \times 1.21 \times 10^{- 5} \times 20}$

T = 40 1.21

n = $\frac{1}{2 \times 1}\sqrt{\frac{40 \times 1.21}{0.1}} = 11Hz$