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Waves question

2025 · 7 Apr · Shift 1 · Q56
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Waves question

2025 · 7 Apr · Shift 1 · Q56

JEE MainPhysicsWavesMCQ+4 / −1
Two harmonic waves moving in the same direction superimpose to form a wave x=acos⁡(1.5t)cos⁡(50.5t)x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})x=acos(1.5t)cos(50.5t) where t is in seconds. Find the period with which they beat. (close to nearest integer)
  1. A
    1 s
  2. B
    4 s
  3. C
    2 s
  4. D
    6 s
View written solutionFree

Correct answer: C

  1. Identify the beat form

The resultant wave is given as x=acos⁡(1.5t)cos⁡(50.5t).x=a\cos(1.5t)\cos(50.5t).x=acos(1.5t)cos(50.5t).

For two harmonic waves of nearly equal angular frequencies ω1\omega_1ω1​ and ω2\omega_2ω2​ moving in the same direction, x=2Acos⁡(ω1−ω22t)cos⁡(ω1+ω22t).x=2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\cos\left(\frac{\omega_1+\omega_2}{2}t\right).x=2Acos(2ω1​−ω2​​t)cos(2ω1​+ω2​​t).

Comparing with x=acos⁡(1.5t)cos⁡(50.5t),x=a\cos(1.5t)\cos(50.5t),x=acos(1.5t)cos(50.5t), we get ω1−ω22=1.5.\frac{\omega_1-\omega_2}{2}=1.5.2ω1​−ω2​​=1.5.

So, ω1−ω2=3 rad/s.\omega_1-\omega_2=3\ \text{rad/s}.ω1​−ω2​=3 rad/s.

  1. Find the beat angular frequency

The amplitude varies as cos⁡(1.5t).\cos(1.5t).cos(1.5t).

Beats are heard/intensity maxima repeat with beat angular frequency ωb=ω1−ω2=3 rad/s.\omega_b=\omega_1-\omega_2=3\ \text{rad/s}.ωb​=ω1​−ω2​=3 rad/s.

Hence the beat frequency is fb=ωb2π=32π Hz.f_b=\frac{\omega_b}{2\pi}=\frac{3}{2\pi}\ \text{Hz}.fb​=2πωb​​=2π3​ Hz.

  1. Find beat period

Beat period is Tb=1fb=2π3 s.T_b=\frac{1}{f_b}=\frac{2\pi}{3}\ \text{s}.Tb​=fb​1​=32π​ s.

Numerically, Tb≈6.2833≈2.09 s.T_b\approx \frac{6.283}{3}\approx 2.09\ \text{s}. Tb​≈36.283​≈2.09 s.

Nearest integer: Tb≈2 s.T_b\approx 2\ \text{s}. Tb​≈2 s.

  1. Choose the correct option

So the correct option is:

C: 2 s\boxed{\text{C: }2\text{ s}}C: 2 s​

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