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:
- (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 $k=\frac{2 \pi}{\lambda}$
$\therefore \quad \lambda=2 \pi A$
