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Waves question

2025 · 4 Apr · Shift 2 · Q53
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Waves question

2025 · 4 Apr · Shift 2 · Q53

JEE MainPhysicsWavesMCQ+4 / −1
Displacement of a wave is expressed as x(t)=5cos⁡(628t+π2)mx(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) \mathrm{m}x(t)=5cos(628t+2π​)m. The wavelength of the wave when its velocity is 300 m/s300 \mathrm{~m} / \mathrm{s}300 m/s is : (π=3.14)(\pi=3.14)(π=3.14)
  1. A
    0.33 m
  2. B
    0.5 m
  3. C
    3 m
  4. D
    5 m
View written solutionFree

Correct answer: C

  1. The given displacement is x(t)=5cos⁡(628t+π2) mx(t)=5\cos\left(628t+\frac{\pi}{2}\right)\,\text{m}x(t)=5cos(628t+2π​)m

  2. Compare this with the standard form of SHM/wave particle displacement: x(t)=Acos⁡(ωt+ϕ)x(t)=A\cos(\omega t+\phi)x(t)=Acos(ωt+ϕ) So, the angular frequency is ω=628 rad/s\omega=628\ \text{rad/s}ω=628 rad/s

  3. Frequency is related to angular frequency by ω=2πf\omega=2\pi fω=2πf Hence, f=ω2π=6282×3.14f=\frac{\omega}{2\pi}=\frac{628}{2\times 3.14}f=2πω​=2×3.14628​

  4. Calculate the frequency: f=6286.28=100 Hzf=\frac{628}{6.28}=100\ \text{Hz}f=6.28628​=100 Hz

  5. Wave speed, frequency, and wavelength are related by v=fλv=f\lambdav=fλ Therefore, λ=vf=300100=3 m\lambda=\frac{v}{f}=\frac{300}{100}=3\ \text{m}λ=fv​=100300​=3 m

  6. So the wavelength is 3 m\boxed{3\ \text{m}}3 m​

  7. Checking options:

    • A: 0.33 m0.33\,\text{m}0.33m ❌
    • B: 0.5 m0.5\,\text{m}0.5m ❌
    • C: 3 m3\,\text{m}3m ✅
    • D: 5 m5\,\text{m}5m ❌

Therefore, the correct option is C.

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