For the travelling harmonic wave $y(x, t) = 2.0 \cos 2 \pi(10 t - 0.0080 x + 0.35)$,where $x$ and $y$ are in $cm$ and $t$ is in $s$. Calculate the phase difference between oscillatory motion of two points separated by a distance of:
$(a)$ $4 \, m$
$(b)$ $0.5 \, m$
$(c)$ $\lambda / 2$
$(d)$ $3 \lambda / 4$

(N/A) The equation for a travelling harmonic wave is given by $y(x, t) = 2.0 \cos 2 \pi(10 t - 0.0080 x + 0.35)$.
Comparing this with the standard wave equation $y = a \cos(2 \pi \nu t - kx + \phi_0)$,we get:
Propagation constant $k = 0.0160 \pi \, rad/cm$.
The phase difference $\Delta \phi$ between two points separated by distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
$(a)$ For $\Delta x = 4 \, m = 400 \, cm$:
$\Delta \phi = (0.0160 \pi \, rad/cm) \times (400 \, cm) = 6.4 \pi \, rad$.
$(b)$ For $\Delta x = 0.5 \, m = 50 \, cm$:
$\Delta \phi = (0.0160 \pi \, rad/cm) \times (50 \, cm) = 0.8 \pi \, rad$.
$(c)$ For $\Delta x = \lambda / 2$:
Since $k = 2 \pi / \lambda$,$\Delta \phi = (2 \pi / \lambda) \times (\lambda / 2) = \pi \, rad$.
$(d)$ For $\Delta x = 3 \lambda / 4$:
$\Delta \phi = (2 \pi / \lambda) \times (3 \lambda / 4) = 1.5 \pi \, rad$.