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Waves question

2022 · 24 Jun · Shift 2 · Q66
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Waves question

2022 · 24 Jun · Shift 2 · Q66

JEE MainPhysicsWavesNumerical+4 / −1
Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by y=(10cos⁡πxsin⁡2πtT)y = (10\cos \pi x\sin {{2\pi t} \over T})y=(10cosπxsinT2πt​) cm The amplitude of the particle at x=43x = {4 \over 3}x=34​ cm will be ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
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Correct answer: 5

  1. The stationary wave is given by y=10cos⁡(πx)sin⁡(2πtT) cmy = 10\cos(\pi x)\sin\left(\frac{2\pi t}{T}\right) \text{ cm}y=10cos(πx)sin(T2πt​) cm

For a stationary wave of the form y=A(x)sin⁡(2πtT),y = A(x)\sin\left(\frac{2\pi t}{T}\right),y=A(x)sin(T2πt​), the amplitude of the particle at position xxx is the coefficient of the time-dependent sine term in magnitude: a(x)=∣10cos⁡(πx)∣.a(x) = |10\cos(\pi x)|.a(x)=∣10cos(πx)∣.

  1. Now substitute x=43 cm.x = \frac{4}{3} \text{ cm}.x=34​ cm. Then a=∣10cos⁡(π⋅43)∣=∣10cos⁡(4π3)∣.a = \left|10\cos\left(\pi \cdot \frac{4}{3}\right)\right| = \left|10\cos\left(\frac{4\pi}{3}\right)\right|.a=​10cos(π⋅34​)​=​10cos(34π​)​.

  2. Evaluate the cosine: cos⁡(4π3)=−12.\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}.cos(34π​)=−21​. So, a=∣10(−12)∣=5 cm.a = \left|10\left(-\frac{1}{2}\right)\right| = 5 \text{ cm}.a=​10(−21​)​=5 cm.

  3. Therefore, the amplitude of the particle at x=43x=\frac{4}{3}x=34​ cm is 5.\boxed{5}.5​.

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