JEE Main 2021 · 17 March · evening · Single correct · +4 / −1

A sound wave of frequency 245 Hz travels with the speed of 300 ms⁻¹ along the positive x-axis. Each point of the wave moves to and from through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

  1. (a) Y(x, t) = 0.03 [ sin 5.1x - (0.2 × 10³)t ]
  2. (b) Y(x, t) = 0.03 [ sin 5.1x - (1.5 × 10³)t ]
  3. (c) Y(x, t) = 0.06 [ sin 5.1x - (1.5 × 10³)t ]
  4. (d) Y(x, t) = 0.06 [ sin 0.8x - (0.5 × 10³)t ]
Answer and solution

Answer (b) Y(x, t) = 0.03 [ sin 5.1x - (1.5 × 10³)t ]

Solution

Physics sir · 3 steps

Read off three numbers: amplitude from the 6 cm swing, ω from the frequency, k from ω/v.

  1. Amplitude. A point moving to and fro covers the full distance from one extreme to the other, which is twice the amplitude. So A = 6/2 = 3 cm = 0.03 m.
  2. Angular frequency. ω = 2π f = 2π × 245 = 1539 rad s⁻¹, which is about 1.5 × 10³ rad s⁻¹.

    k = ω/v = 1539/300 ≈ 5.1 m⁻¹

  3. So. A wave moving along +x has the form y = A sin(kx - ω t), giving y = 0.03 sin(5.1x - 1.5 × 10³ t). That is option (b).

The idea to keep. The total to-and-fro distance is 2A, not A. And both ω and k carry a 2π: ω = 2π f, k = 2π/λ = ω/v.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Mechanical Waves

Asked about 60 times in JEE Main since 2020. This question is from 2021.