Wave Basics & Properties of Sound question from JEE Main 2013 (25 Apr, Shift 1)

Answer: (b)

1. Crest to neighbouring trough at one instant is $\frac{\lambda}{2} = 4$ cm, so $\lambda = 8$ cm.
2. Crest to trough at the same place (vertical) is $2A = 1$ cm, so $A = 0.5$ cm.
3. Next crest at the same place after one period: $T = 0.4$ s, so $\omega = \frac{2\pi}{T} = 5\pi$ rad/s.
4. $v_{max} = A\omega = 0.5 \times 5\pi = \frac{5\pi}{2}$ cm/s
5. So the answer is (b) $\frac{5\pi}{2}$ cm/s.