For the harmonic travelling wave $y=2 \cos 2 \pi(10 t-0.0080 x+3.5)$ where $x$ and $y$ are in cm and $t$ is in second. 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 $x=100 \mathrm{~cm}$, at $t=T \mathrm{sec}$ and $t=5$ ?
$$\begin{aligned} &\text { Given, wave functions are }\\ &\begin{aligned} y & =2 \cos 2 \pi(10 t-0.0080 x+3.5) \\ & =2 \cos (20 \pi t-0.016 \pi x+7 \pi) \end{aligned} \end{aligned}$$
Now, standard equation of a travelling wave can be written as
$$y=a \cos (\omega t-k x+\phi)$$
On comparing with above equation, we get
$$\begin{aligned} & a=2 \mathrm{~cm} \\ & \omega=20 \pi \mathrm{rad} / \mathrm{s} \\ & k=0.016 \pi \end{aligned}$$
Path difference = 4 cm
$$ \begin{aligned} &\text { (a) Phase difference } \Delta \phi=\frac{2 \pi}{\lambda} \times \text { Path difference }\\ &\begin{aligned} \therefore \quad \Delta \phi & =0.016 \pi \times 4 \times 100 \quad \left(\because \frac{2 \pi}{\lambda}=k\right)\\ & =6.4 ~\pi \mathrm{~rad} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { (b) } \Delta \phi=\frac{2 \pi}{\lambda} \times(0.5 \times 100) \quad [\because \text { Path difference }=0.5 \mathrm{~m}] \\ =0.016 ~\pi \times 0.5 \times 100 \\ =0.8 \pi ~\mathrm{~rad} \end{aligned}$$
(c) $\Delta \phi=\frac{2 \pi}{\lambda} \times\left(\frac{\lambda}{2}\right)=\pi \mathrm{rad}\quad$ [$$\because$$ Path difference = $$\lambda/2$$]
(d) $\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}=\frac{3 \pi}{2} \mathrm{rad}$
(e) $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{20 \pi}=\frac{1}{10} \mathrm{~s}$
$$\begin{aligned} \therefore \text { At } \quad x & =100 \mathrm{~cm} \\ t & =T \\ \phi_1 & =20 \pi T-0.016 \pi(100)+7 \pi \\ & =20 \pi\left(\frac{1}{10}\right)-1.6 \pi+7 \pi=2 \pi-1.6 \pi+7 \pi \quad \text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\text { Again, at } x=100 \mathrm{~cm}, t=5 \mathrm{~s}\\ &\begin{aligned} \phi_2 & =20 \pi(5)-0.016 \pi(100)+7 \pi \\ & =100 \pi-(0.016 \times 100) \pi+7 \pi \\ & =100 \pi-1.6 \pi+7 \pi \quad \text{... (ii)} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\therefore \text { From Eqs. (i) and (ii), we get }\\ &\begin{aligned} \Delta \phi=\text { phase difference } & =\phi_2-\phi_1 \\ & =(100 \pi-1.6 \pi+7 \pi)-(2 \pi-1.6 \pi+7 \pi) \\ & =100 \pi-2 \pi=98 \pi \mathrm{rad} \end{aligned} \end{aligned}$$