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Stationary Waves & Vibrating Strings question from JEE Main 2013

Q105JEE Main 2013Difficult

Stationary Waves & Vibrating Strings

A sonometer wire of length 1.5 m1.5\text{ m} is made of steel. The tension in it produces an elastic strain of 1%1\%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7×103 kg/m37.7 \times 10^{3}\text{ kg/m}^{3} and 2.2×1011 N/m22.2 \times 10^{11}\text{ N/m}^{2} respectively?

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Answer: (b)

  1. Stress TA=Y×strain=2.2×1011×0.01=2.2×109 N/m2\frac{T}{A}=Y\times\text{strain}=2.2\times10^{11}\times0.01=2.2\times10^{9}\text{ N/m}^2.
  2. v=TAρ=2.2×1097.7×103=2.857×105≈534.5 m/sv=\sqrt{\frac{T}{A\rho}}=\sqrt{\frac{2.2\times10^9}{7.7\times10^3}}=\sqrt{2.857\times10^5}\approx534.5\text{ m/s}
  3. n=v2L=534.53≈178.2 Hzn=\frac{v}{2L}=\frac{534.5}{3}\approx178.2\text{ Hz}
  4. So the answer is (b).
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