Two particles oscillate in simple harmonic motion along a common straight-line segment of length $\mathit{A}$. Each particle has a period of $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{s}}$ , but they differ in phase by $\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{\text{rad}}$.

1. How far apart are they (in terms of$\mathit{A}$ ) $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{s}}$after the lagging particle leaves one end of the path?
2. Are they then moving in the same direction, toward each other, or away from each other?

1. Period of oscillation, $\text{T}=1.5\text{s}$
2. Phase difference of the motion, $\varphi =\frac{\mathrm{\pi }}{6}$
3. Time $t=0.50$ (to be used for finding the separation).

Using the equation of position, we can find the separation between two particles to be$\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{s}}$after the lagging particle leaves one end of the path. Similarly, using the equations of velocity, we can determine whether they are moving in the same direction, towards each other, or away from each other.

Formulae:

The general expression for velocity of motion,

$x=x_m\cos \left(\omega t+\varphi \right)$ ......(i)

The general expression for velocity of motion,

$v=-x_m\omega \sin \left(\omega t+\varphi \right)$ ......(ii)

For particle 1 and particle 2, the general expression of motion can be given as,

$$\begin{gathered} x_1=x_m\cos \left(\omega t\right) \\ {x}_2=x_m\cos (\omega t+\varphi ) \end{gathered}$$

The amplitude of both particles is $x_m=\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$because the range of the motion is

$$\begin{gathered} x_1=\frac{A}{2}\cos \left(\frac{2\pi t}{T}\right) \\ {x}_2=\frac{A}{2}\cos \left(\frac{2\pi t}{T}+\frac{\pi }{6}\right) \end{gathered}$$

Particle1 is at one end when $t=0$and Particle2 is at $\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$when,

$$\begin{gathered} \left(\frac{2\pi t}{T}+\frac{\pi }{6}\right)=0 \\ t=-T/12 \end{gathered}$$

Therefore, Particle1 lags Particle2 by $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\text{th}$of the period.

As $\text{t}=0.50\text{s},\text{T}=1.5\text{s}$

$$\begin{gathered} x_1=\frac{A}{2}\cos \left(\frac{2\pi \times 0.50s}{1.5s}\right) \\ =-0.25A...................................\text{(i)} \end{gathered}$$

$\begin{array}{rcl}{x}_2& =& \frac{A}{2}\cos \left(\frac{2\pi \times 0.50s}{1.5s}+\frac{\pi }{6}\right)\\ & =& -0.43A....................................\text{(ii)}\end{array}$

Therefore, the separation between them at time $t=0.50\text{s}$using equations (i) & (ii) is given as

$\begin{array}{rcl}{x}_1-x_2& =& -0.25A+0.43A\\ & =& 0.18A\end{array}$

Therefore, the separation between two particles after the lagging particle leaves one end of the path is$0.18\text{A}$

For velocity of particles,

$\begin{array}{rcl}{v}_1& =& \frac{d{x}_1}{dt}\\ & =& \frac{d}{dt}\left[\frac{A}{2}\cos \left(\frac{2\pi t}{T}\right)\right]\\ & =& -\frac{\pi A}{T}\sin \left(\frac{2\pi t}{T}\right)\end{array}$

$\begin{array}{rcl}{v}_2& =& \frac{d{x}_2}{dt}\\ & =& \frac{d}{dt}\left[\frac{A}{2}\cos \left(\frac{2\pi t}{T}+\frac{\pi }{6}\right)\right]\\ & =& -\frac{\pi A}{T}\sin \left(\frac{2\pi t}{T}+\frac{\pi }{6}\right)\end{array}$

As, $\text{t}=0.50\text{s},\text{T}=1.5\text{s}$

$\begin{array}{rcl}{\text{v}}_1& =& -\frac{\pi \text{A}}{1.5\text{s}}\text{sin}\left(\frac{2\pi \times 0.50\text{s}}{1.5\text{s}}\right)\\ & =& -0.076\text{A}\end{array}$

$\begin{array}{rcl}{\text{v}}_2& =& -\frac{\pi \text{A}}{1.5\text{s}}\text{sin}\left(\frac{2\pi \times 0.50\text{s}}{1.5\text{s}}+\frac{\pi }{6}\right)\\ & =& -0.0956\text{A}\end{array}$

As the value of both the velocities is negative, it indicates that the particles are moving in the same direction.