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MHT CET 2024 (15 May, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from MHT CET 2024 (15 May, Shift 1)

Q47MHT CET 2024 · 15 MAY · SHIFT 1Medium

Stationary Waves & Vibrating Strings

Two progressive waves Y1=sin⁡2π(t0.4−x4)Y_1 = \sin 2\pi\left(\frac{t}{0.4} - \frac{x}{4}\right) and Y2=sin⁡2π(t0.4+x4)Y_2 = \sin 2\pi\left(\frac{t}{0.4} + \frac{x}{4}\right) superpose to form a standing wave. ' xx ' and ' yy ' are in SI system. Amplitude of the particle at x=0.5 mx = 0.5\text{ m} is [sin⁡45∘=cos⁡45∘=12]\left[\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\right]

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Answer: (c)

  1. sin⁡(A−B)+sin⁡(A+B)=2sin⁡Acos⁡B\sin(A - B) + \sin(A + B) = 2\sin A\cos B, so Y=2cos⁡(2πx4)sin⁡(2πt0.4)Y = 2\cos\left(\dfrac{2\pi x}{4}\right)\sin\left(\dfrac{2\pi t}{0.4}\right).
  2. Amplitude at xx: A(x)=2cos⁡(πx2)A(x) = 2\cos\left(\dfrac{\pi x}{2}\right).
  3. At x=0.5 mx = 0.5\text{ m}: A=2cos⁡45∘=2×12=2 mA = 2\cos 45^\circ = 2 \times \dfrac{1}{\sqrt{2}} = \sqrt{2}\text{ m}.
  4. So the answer is (c) 2 m\sqrt{2}\text{ m}.
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