The equation of a wave is $y(x,t)=0.05\sin\left(\frac{\pi}{2}(10x-40t)-\frac{\pi}{4}\right)\ \mathrm{m}$…
The equation of a wave is $y(x,t)=0.05\sin\left(\frac{\pi}{2}(10x-40t)-\frac{\pi}{4}\right)\ \mathrm{m}$
Find
(i) the wavelength, frequency and wave velocity,
(ii) the particle velocity and acceleration at $x=0.5\ \mathrm{m}$ and $t=0.05\ \mathrm{s}$.
Solution
Sol. (i) The given equation may be rewritten as,
$y(x,t)=0.05\sin(5\pi x-20\pi t-\pi/4)\ \mathrm{m}$
Comparing this with standard equation of plane progressive harmonic wave,
$y(x,t)=A\sin(kx-\omega t+\phi)$, we get
Wave number, $k=\dfrac{2\pi}{\lambda}=5\pi\ \mathrm{rad\ m^{-1}}$
$\therefore\ \lambda=0.4\ \mathrm{m}$
Angular frequency,
$\omega=2\pi f=20\pi\ \mathrm{rad\ s^{-1}}\ \Rightarrow\ f=10\ \mathrm{Hz}$
Wave velocity,
$v=f\lambda=\dfrac{\omega}{k}=4\ \mathrm{ms^{-1}}$ in + x-direction
(ii) The particle velocity and acceleration at $x=0.5\ \mathrm{m}$ and $t=0.05\ \mathrm{s}$ are
$\dfrac{dy}{dt}=-(20\pi)(0.05)\cos\left(\dfrac{5\pi}{2}-\pi-\dfrac{\pi}{4}\right)=2.22\ \mathrm{ms^{-1}}$
$\dfrac{d^{2}y}{dt^{2}}=-(20\pi)^{2}(0.05)\sin\left(\dfrac{5\pi}{2}-\pi-\dfrac{\pi}{4}\right)=140\ \mathrm{ms^{-2}}$
Answer: 140 ms$^{-2}$