Question

A travelling harmonic wave on a string is described by
$y(x,t)=7.5\sin (0.0050x+12t+\pi /4)$
(a) What are the displacement and velocity of oscillation of a point at $x=1$ cm and $t=1$ s? Is this velocity equal to the velocity of wave propagation ?
(b) Locate the points of the string which have the same transverse displacements and velocity as the $x=1$ cm point at $t=2s$ , 5 s and 11s

Answer 1

(a) The given harmonic wave is

$y(x,t)=7.5sin(0.0050x+12t+\frac{\pi }{4})$

For $x=1cm$ and $t=1s$,

$y(1,1)=7.5sin(0.0050x+12+\frac{\pi }{4})$

$=7.5sin(12.0050+\frac{\pi }{4})$

$=7.5sin\theta$

Where,

$\theta =12.0050+\frac{\pi }{4}=12.0050+\frac{3.14}{4}=12.79rad$

$=\frac{180}{3.14}\times 12.79={732.81}^o$

$\therefore y=(1,1)=7.5sin\left({732.81}^o\right)$

$=7.5sin(90\times 8+{12.81}^o)=7.5sin{12.81}^o$

$=7.5\times 0.2217$

$=1.6629\approx 1.663cm$

The velocity of the oscillation at a given point and times is given as:

$v=\frac{d}{dt}y(x,t)=\frac{d}{dt}\left[7.5sin(0.0050x+12t+\frac{\pi }{4})\right]$

$=7.5\times 12cos(0.0050x+12t+\frac{\pi }{4})$

At $x=1cm$ and $t=1s$

$v=y(1,1)=90cos(12.005+\frac{\pi }{4})$

$=90coss\left({732.81}^o\right)=90cos(90\times 8+{12.81}^o)$

$=90cos\left({12.81}^o\right)$

$=90\times 0.975=87.75cm/s$

Now, the equation of a propagating wave is given by:

$y(x,t)=asin(kx+wt+\varphi )$

where,

$k=2\pi /\lambda$

$\therefore \lambda =2\pi /k$

and, $\omega =2\pi v$

$\therefore v=\omega /2\pi$

Speed, $v=v\lambda =\omega /k$

Where,

$\omega =12rad/s$

$k=0.0050m^{-1}$

$\therefore v=12/0.0050=2400cm/s$
Hence, the velocity of the wave oscillation at $x=1cm$ and $t=1s$ is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:

$k=2\pi /\lambda$

$\therefore \lambda =2\pi /k=2\times 3.14/0.0050$

$=1256cm=12.56m$
Therefore, all the points at distance $n\lambda (n=\pm 1\pm 2,...$ and so on$)$, i.e. $\pm 12.56m,\pm 25.12m,...$ and so on for $x=1cm$, will have the same displacement as the $x=1cm$ points at $t=2s,5s$ and $11s$.