JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 15)
Two particles are performing simple harmonic motion in a straight line about
the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has
displacement A while the other one has displacement $${{ - A} \over 2}$$ and they are moving towards each other. If they cross each other at time t, then t is :
$${T \over 6}$$
$${5T \over 6}$$
$${T \over 3}$$
$${T \over 4}$$
Explanation
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Angular displacement ($$\theta $$1) of particle 1. from equilibrium,
$${y_1}$$ = A sin$$\theta$$1
$$ \Rightarrow $$ A = Asin$$\theta $$1
$$ \Rightarrow $$ sin$$\theta $$1 = 1 = sin $${\pi \over 2}$$
$$ \therefore $$ $$\theta $$1 = $${\pi \over 2}$$
Similarly for particle 2 angular displacement $$\theta $$2 from equilibrium,
y2 = Asin$$\theta $$2
$$ \Rightarrow $$ $$-$$ $${A \over 2}$$ = Asin$$\theta $$2
$$ \Rightarrow $$ sin$$\theta $$2 = $$-$$ $${1 \over 2}$$ = sin$$\left( { - {\pi \over 3}} \right)$$
$$ \Rightarrow $$ $$\theta $$2 = $$-$$ $${{\pi \over 3}}$$
Relative angular displacement of the two particle,
$$\theta $$ = $$\theta $$1 $$-$$ $$\theta $$2
= $${{\pi \over 2}}$$ $$-$$ $$\left( { - {\pi \over 6}} \right)$$
= $${{{2\pi } \over 3}}$$
Relative angular velocity $$=$$ $$\omega - \left( { - \omega } \right)$$ = $$2\omega $$
If they cross each other at time t
then, t = $${\theta \over {2\omega }}$$ = $${{2\pi } \over {3 \times 2\omega }}$$ = $${\pi \over {3 \times {{2\pi } \over T}}}$$ = $${T \over 6}$$
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