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JEE Main 2019 (9 Apr, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from JEE Main 2019 (9 Apr, Shift 1)

Q93JEE Main 2019 · 9 APR · SHIFT 1Medium

Stationary Waves & Vibrating Strings

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin(0.157x) cos(200π\pit). The length of the string is: (All quantities are in SI units.)

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Answer: (c)

  1. k=0.157=π20 m−1k = 0.157 = \dfrac{\pi}{20}\text{ m}^{-1}, so λ=2πk=40 m\lambda = \dfrac{2\pi}{k} = 40\text{ m}.
  2. 4th harmonic: L=4×λ2=2λL = 4 \times \dfrac{\lambda}{2} = 2\lambda.
  3. L=2×40=80 mL = 2 \times 40 = 80\text{ m}.
  4. So the answer is (c) 80 m80\text{ m}.
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