Question

The equation of a wave is, y(x,t) = 0.05 $sin[\frac{\pi }{2}(10x-40t)-\frac{\pi }{4}]m$. Find:
(a) The wavelength, the frequency and the wave velocity
(b) The particle velocity and acceleration at x = 0.5 m and t = 0.05 s.

Answer 1

Wave equation $y(x,t)=0.05\sin \left[\frac{\pi }{2}\right(10x-40t)-\frac{\pi }{4}]$

So, $y(x,t)=0.05\sin \left[\right(5\pi x-20\pi t)-\frac{\pi }{4}]$

Comparing with $y(x,t)=A\sin [kx-wt+\varphi ]$

We get, $w=20\pi$ $k=5\pi$ $A=0.05m$

Wavelength $\lambda =\frac{2\pi }{k}=\frac{2\pi }{5\pi }=0.4m$

Frequency $f=\frac{w}{2\pi }=\frac{20\pi }{2\pi }=10Hz$

Wave velocity $v=f\lambda =10\times 0.4=4m/s$

Particle velocity $V_p=\frac{dy}{dt}=0.05\times (-20\pi )\cos [5\pi x-20\pi t-\frac{\pi }{4}]$

At $x=0.5m,t=0.05s$, $V_p=0.05\times (-20\pi )\cos \left[5\pi \right(0.5)-20\pi (0.05)-\frac{\pi }{4}]$

Or, $V_p=-20\pi \times 0.05\cos \left(\frac{5\pi }{4}\right)=-20\pi \times 0.05\times (-\frac{1}{\sqrt{2}})=2.22m/s$

Particle acceleration $a_p=\frac{d{V}_p}{dt}=0.05\times (-20\pi )(-20\pi )(-\sin [5\pi x-20\pi t-\frac{\pi }{4}\left]\right)$

At $x=0.5m,t=0.05s$, $a_p=-0.05\times \left(20\pi {)}^2\sin \right[5\pi \left(0.5\right)-20\pi \left(0.05\right)-\frac{\pi }{4}]$

Or, $a_p=-(20\pi {)}^2\times 0.05\sin (\frac{5\pi }{4})=-(20\pi {)}^2\times 0.05\times (-\frac{1}{\sqrt{2}})=139.6m/s^2$