Question

Two particles are executing identical simple harmonic motions described by the equations, $x_1=a\cos (\omega t+\left(\frac{\pi }{6}\right))$ and $x_2=a\cos (\omega t+\frac{\pi }{3})$. The minimum interval of time between the particles crossing the respective mean position is?

• A. $\frac{\pi }{2\omega }$
• B. $\frac{\pi }{3\omega }$
• C. $\frac{\pi }{4\omega }$
• D. $\frac{\pi }{6\omega }$

Answer 1

The correct option is D $\frac{\pi }{6\omega }$

Equations are $x_1=a\cos (\omega t+\frac{\pi }{6})$
and $x_2=a\cos (\omega t+\frac{\pi }{6})$
The first will pass through the mean position when $x_1=0$ i.e., for instants $t$ for which $(\omega t+\frac{\pi }{6})=\frac{n\pi }{2}$, where $n$ is an integer.
The smallest value for $t$ is $n=1$

$\omega t_1=\left(\pi /2\right)-\left(\pi /6\right)=\pi /3$.
The second will pass through the mean position when $x_2=0$ i.e., for instants $t$ for which $(\omega t+\frac{\pi }{3})=\frac{m\pi }{2}$
where $m$ is an integer.

The smallest value for $t$ is $m=1$

$\omega t_2=\left(\pi /2\right)-\left(\pi /3\right)=\pi /6$

The smallest interval between the instants $x_1=0$ and $x_2=0$ therefore,

$\omega (t_1-t_2)=(\frac{\pi }{3}-\frac{\pi }{6})=\frac{\pi }{6}\Rightarrow t_1-t_2=\frac{\pi }{6\omega }$