Question

Can you please answer quesion 6 part a, b, c and d
NAME Last 4 digits ID 6. (20 points total) A transverse wave on a string is described by the wave function y(x,t) = 0.120 sin( (π x / 8) + 4 where x an y are in meters and t is in seconds. (a) (4 points) What is the wavelength for this wave? meters wavelength = (b) (4 points) What is the period of this wave? seconds Period = (4 points) What is the speed of propagation of this wave? (c) meters/second Speed =

Answer 1

We know that the general equation of a transverse wave is given as

$y(x,t)=Asin(\omega t+kx)$

Comparing this equation with the given equation we find,

$\omega=4\pi \ and \ k=\frac{\pi}{8}$

Now, we also know that

$\\ k=\frac{2\pi}{\lambda} \Rightarrow \frac{\pi}{8}=\frac{2\pi}{\lambda} \\ or, \lambda \ i.e \ wavelength=16 \ m \ (ans \ a)$

$\\ \omega=\frac{2\pi}{T} \Rightarrow 4\pi=\frac{2\pi}{T} \\ or, T \ i.e \ Time \ period=0.5 \ s \ (ans \ b)$

Now, speed of propagation of a transvers wave is equal to

$v=f\lambda \ ; also, v=\frac{\omega}{k}\Rightarrow \frac{4\pi}{\frac{\pi}{8}}=32 \ m/s \ (ans \ c)$

transvere speed of a particle associated to this wave, i.e particle velocity is given as

$\\ \frac{dy}{dt}\Rightarrow v=4\pi\left (0.12cos(\frac{\pi x}{8}+4\pi t) \right ) \\ at \ x=1.6 \ m \ and \ t=0.2 \ s$

$\\ v=4\pi\left (0.12cos(\frac{\pi 1.6}{8}+4\pi 0.2) \right ) \\ v=1.507cos(\pi) \\or, v=-1.507 \ m/s \ (ans \ d)$