Waves - Result Question 4

4. A transverse wave is represented by $y=A \sin$ $(\omega t-k x)$. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

[2010]

(a) $\frac{\pi A}{2}$

(b) $\pi A$

(c) $2 \pi A$

(d) $A$

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Answer:

Correct Answer: 4. (c)

Solution:

  1. (c) $y=A \sin (\omega t-k x)$

Particle velocity,

$v_p=\frac{d y}{d t}=A \omega \cos (\omega t-k x)$

$\therefore \quad v _{p \max }=A \omega$

wave velocity $=\frac{\omega}{k}$

$\therefore A \omega=\frac{\omega}{k}$

i.e., $A=\frac{1}{k}$ But $k=\frac{2 \pi}{\lambda}$

$\therefore \quad \lambda=2 \pi A$