Question

A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of $3m/s$ and $90m/s^2$ respectively. Velocity of the wave is $20m/s$. Find the waveform.

• A. $y=\left(10cm\right)\sin (30t\pm \frac{3}{2}x+\varphi )$
• B. $y=\left(20cm\right)\sin (30t\pm \frac{3}{2}x+\varphi )$
• C. $y=\left(10cm\right)\sin (60t\pm \frac{3}{2}x+\varphi )$
• D. $y=\left(20cm\right)\sin (60t\pm \frac{3}{2}x+\varphi )$

Answer 1

The correct option is A $y=\left(10cm\right)\sin (30t\pm \frac{3}{2}x+\varphi )$

Let the equation of wave be $y(x,t)=A\sin (kx\pm \omega t+\varphi )$

we use $\pm$ here as no information is given about the direction of wave.

at any position $x=x_o$, the velocity is given as

$v\left(t\right)=dy(x_o,t)/dt=A\omega \cos (k{x}_o-\omega t+\varphi )$

and acceleration is given as

$a\left(t\right)=dv\left(t\right)/dt=-A{\omega }^2\sin (k{x}_o-\omega t+\varphi )$

The ratio of magnitude of maximum velocity to maximum acceleration is

$1/\omega =3/90\Rightarrow \omega =30$

So maximum velocity magnitude is

$A\omega =3\Rightarrow A\times 30=3\Rightarrow A=0.1m=10cm$

Now speed of wave is $v=\nu \times \lambda =\omega /k=20m/s$

$\Rightarrow k=30/20=3/2$

So equation of wave is written as: $y(x,t)=\left(10cm\right)\sin (\frac{3}{2}x\pm 30t+\varphi )$