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JEE MAIN - Physics (2011 - No. 15)

Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega $$ along the $$x$$-axis. Their mean position is separated by distance $${X_0}\left( {{X_0} > A} \right)$$. If the maximum separation between them is $$\left( {{X_0} + A} \right),$$ the phase difference between their motion is:
$${\pi \over 3}$$
$${\pi \over 4}$$
$${\pi \over 6}$$
$${\pi \over 2}$$

Explanation

We know that, equation for SHM along x-axis is given by $$x = A\sin (\omega t + \phi )$$

Let mean position for 1st particle is at x = 0

So, the SHM equation for 1st particle,

$${x_1} = A\sin (\omega t + {\phi _1})$$

Now, as the separation between mean positions of both the particle is x$_0$

So, mean position for 2nd particle is $x=x_0$

Hence, the SHM equation for 2nd particle,

$${x_2} = {x_0} + A\sin (\omega t + {\phi _2})$$

Now, the separation between particles would be

$$\left| {{x_2} - {x_1}} \right| = {x_0} + A\sin (\omega t + {\phi _2}) - A\sin (\omega t + {\phi _1})$$

$$ = {x_0} + A\left[ {\sin (\omega t + {\phi _2}) - \sin (\omega t + {\phi _1})} \right]$$

We know,

$$\sin C - \sin D = 2\cos \left( {{{C + D} \over 2}} \right)\sin \left( {{{C - D} \over 2}} \right)$$

Hence,

$$\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right)\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)$$

For maximum separation,

Let $$\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right) = 1$$

So, $$\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)$$

Since, $$\left| {{x_2} - {x_1}} \right| = {x_0} + A$$ (given)

So, $${x_0} + A = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)$$

$$ \Rightarrow \sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right) = {1 \over 2}$$

$$ \Rightarrow {{{\phi _2} - {\phi _1}} \over 2} = {\pi \over 6}$$

$$ \Rightarrow {\phi _2} - {\phi _1} = {\pi \over 3}$$

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