For the harmonic travelling wave $y = 5 \cos 2\pi (10t - 0.008x + 3.5)$ where $x$ and $y$ are in $cm$ and $t$ is in seconds. What is the phase difference between the oscillatory motion at two points separated by a distance of:
$(a)$ $4 \ m$
$(b)$ $0.5 \ m$
$(c)$ $\frac{\lambda}{2}$
$(d)$ $\frac{3\lambda}{4}$ (at a given instant of time)
$(e)$ What is the phase difference between the oscillation of a particle located at $x = 100 \ cm$,at $t = T \ s$ and $t = 5 \ s$?

(A) The given wave equation is $y = 5 \cos(20\pi t - 0.016\pi x + 7\pi)$.
Comparing this with the standard form $y = a \cos(\omega t - kx + \phi)$,we get:
$k = 0.016\pi \ rad/cm$ and $\omega = 20\pi \ rad/s$.
$(a)$ For $\Delta x = 4 \ m = 400 \ cm$,the phase difference is $\Delta \phi = k \Delta x = 0.016\pi \times 400 = 6.4\pi \ rad$.
$(b)$ For $\Delta x = 0.5 \ m = 50 \ cm$,the phase difference is $\Delta \phi = k \Delta x = 0.016\pi \times 50 = 0.8\pi \ rad$.
$(c)$ For $\Delta x = \frac{\lambda}{2}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \ rad$.
$(d)$ For $\Delta x = \frac{3\lambda}{4}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = 1.5\pi \ rad$.
$(e)$ For a particle at a fixed position,the phase difference between two times $t_1$ and $t_2$ is $\Delta \phi = \omega(t_2 - t_1)$.
Given $t_1 = T = \frac{2\pi}{\omega} = \frac{2\pi}{20\pi} = 0.1 \ s$ and $t_2 = 5 \ s$,
$\Delta \phi = 20\pi(5 - 0.1) = 20\pi(4.9) = 98\pi \ rad$.