Question

A progressive simple harmonic wave is moving in air along the $x$-axis. The part of this wave at a given point $x=x_0$ from the source and at a certain instant $t=t_0$ has the waveform shown below in the displacement-time $(y-t)$ graph and velocity-time (v-t) graph respectively. Velocity of the wave has value $V_0$ and its angular frequency is $\omega$. Which of the following equations will correctly represent the complete wave at $x=x_0$, agreeing with above waveforms?

• A. $y=-A\left[\cos \left\{\frac{2\pi }{T}\right(t-t_0)-\frac{\pi }{2}\}\right]$
• B. $y=-A\left[\sin \left\{\frac{2\pi }{T}\right(t-t_0)+\frac{\pi }{2}\}\right]$
• C. $y=A\left[\sin \left\{\frac{2\pi }{T}\right(t-t_0)+\frac{\pi }{2}\}\right]$
• D. $y=A\left[\cos \left\{\frac{2\pi }{T}\right(t-t_0)-\frac{\pi }{2}\}\right]$

Answer 1

The correct option is C $y=A\left[\sin \left\{\frac{2\pi }{T}\right(t-t_0)+\frac{\pi }{2}\}\right]$

Any general simple harmonic progressive wave is expressed as

$y=A_0\sin (\omega t-kx+{\varphi }_0)$

Thus, the particle velocity is given by $v=\frac{dy}{dt}=A_0\omega \cos (\omega t-kx+{\varphi }_0)$

Given that, at $x=x_0$ and $t=t_0$, we have $y=A$ and $v=0$

Thus, $\cos (\omega t_0-k{x}_0+{\varphi }_0)=0\Rightarrow \omega t_0-k{x}_0+{\varphi }_0=\pm \pi /2$

But, for $t<t_0$, $v>0$. Thus, $\omega t_0-k{x}_0+{\varphi }_0=\pi /2$

Hence, $\sin (\omega t_0-k{x}_0+{\varphi }_0)=1\Rightarrow A_0=A$

Hence, ${\varphi }_0=\pi /2-\omega t_0+k{x}_0$

Substituting this in the wave equation,

$y=A\sin (\omega t-kx+\pi /2-\omega t_0+k{x}_0)$

At $x=x_0$, the wave equation is $y=A\sin \left(\omega \right(t-t_0)+\pi /2)$

Replacing $\omega$ by $\frac{2\pi }{T}$

we have $y=A[\sin (\frac{2\pi }{T}(t-t_0)+\frac{\pi }{2})]$