Question

Equation of a progressive wave is given by $y = 0.2 cos \, \pi \big(0.04 t + 0.02 x - \frac{\pi}{6}\big).$ 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

Answer 1

Answer: (C)

25 cm

Answer 2

Comparing with y = a cos $(\omega t + kx - \phi)$ We get, \hspace25mm k = \frac{2 \pi}{\lambda} = 0.02 \pi \Rightarrow \hspace30mm \lambda = 100 \, km Also, it is given that phase difference between particles $? \phi = \frac{\pi}{2}.$ Hence, path difference between them \hspace40mm ? x = \frac{\lambda}{2 \pi} \times ? \phi \hspace40mm = \frac{\lambda}{2 \pi} \times \frac{\pi}{2} \hspace40mm = \frac{\lambda}{4} = \frac{100}{4} \hspace40mm = 25 \, cm