For a wave described by $y = A \sin (\omega t - kx)$, consider the following points (i) $x = 0$ (ii) $x =…
For a wave described by $y = A \sin (\omega t - kx)$, consider the following points
(i) $x = 0$
(ii) $x = \frac{\pi}{4k}$
(iii) $x = \frac{\pi}{2k}$
(iv) $x = \frac{3\pi}{4k}$
For a particle at each of these points at $t = 0$, describe whether the particle is moving or not and in what direction and describe whether the particle is speeding up, slowing down or instantaneously not accelerating.
Solution
Sol. Given, $y = A \sin(\omega t - kx)$
Particle velocity, $v_P (x,t) = \frac{dy}{dt} = \omega A \cos(\omega t - kx)$ and particle
acceleration,
$a_P (x,t) = \frac{d^2 y}{dt^2} = - \omega^2 A \sin(\omega t - kx)$
(i) $t = 0$ and $x = 0$, $v_P = + \omega A$ and $a_P = 0$
i.e. Particle is moving upwards but its acceleration is zero.
Note Direction of velocity can be obtained in a different manner as under,
At $t = 0, y = A \sin(-kx) = - A \sin kx$
(i. e. y-x graph is as shown in figure. At $x = 0$, slope is negative.
Therefore, particle velocity is positive ($v_P = - v \times$ slope) as the wave is travelling along positive x-direction.
(ii) $t = 0, x = \frac{\pi}{4k} \Rightarrow kx = \frac{\pi}{4}$
$v_P = \omega A \cos\left(-\frac{\pi}{4}\right) = + \frac{\omega A}{\sqrt{2}}$
and $a_P = - \omega^2 A \sin\left(-\frac{\pi}{4}\right) = + \frac{\omega^2 A}{\sqrt{2}}$
Velocity of particle is positive, i.e. the particle is moving upwards (along positive y-direction). Further, $v_P$ and $a_P$ are in the same direction (both are positive).
Hence, the particle is speeding up.
(iii) $t = 0, x = \frac{\pi}{2k} \Rightarrow kx = \frac{\pi}{2}$
$v_P = \omega A \cos\left(-\frac{\pi}{2}\right) = 0$
and $a_P = - \omega^2 A \sin\left(-\frac{\pi}{2}\right) = \omega^2 A$
i.e. particle is stationary or at its extreme position (i.e. $y = - A$). So, it is speeding up at this instant.
(iv) $t = 0, x = \frac{3\pi}{4k} \Rightarrow kx = \frac{3\pi}{4}$
$\therefore\; v_P = \omega A \cos\left(-\frac{3\pi}{4}\right) = - \frac{\omega A}{\sqrt{2}}$
$a_P = - \omega^2 A \sin\left(-\frac{3\pi}{4}\right) = + \frac{\omega^2 A}{\sqrt{2}}$
Velocity of particle is negative, i.e. the particle is moving downwards. Further, $v_P$ and $a_P$ are in opposite directions, i.e. the particle is slowing down.
Answer: $+\dfrac{\omega^2 A}{\sqrt{2}}$