The distance is expressed in cm and time in second. What will be the minimum distance between two particles…
The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of $\frac{\pi}{2}$?
(a) 4 cm
(b) 8 cm
(c) 25 cm
(d) 12.5 cm
Solution
Comparing the given equation with $y = a\cos (\omega t + kx - \phi)$,
we get
$k = \frac{2\pi}{\lambda} = 0.02 \Rightarrow \lambda = 100\text{ cm}$
$\Delta\phi = \frac{\pi}{2}$
Hence, path difference between them,
$\Delta x = \frac{\lambda}{2\pi} \times \Delta\phi = \frac{\lambda}{2\pi} \times \frac{\pi}{2} = \frac{\lambda}{4} = \frac{100}{4} = 25\text{ cm}$