Question 145: Oscillations

Question

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 :

Accepted Answer

Angular displacement ($$	heta $$
1
) of particle 1. from equilibrium,
$${y_1}$$ = A sin$$	heta$$
1
$$ \Rightarrow $$   A = Asin$$	heta $$
1
$$ \Rightarrow $$   sin$$	heta $$
1
= 1 = sin $${\pi  \over 2}$$
$$ 	herefore $$   $$	heta $$
1
= $${\pi  \over 2}$$
Similarly for particle 2 angular displacement $$	heta $$
2
from equilibrium,
y
2
= Asin$$	heta $$
2
$$ \Rightarrow $$   $$-$$ $${A \over 2}$$ = Asin$$	heta $$
2
$$ \Rightarrow $$   sin$$	heta $$
2
= $$-$$ $${1 \over 2}$$ = sin$$\left( { - {\pi  \over 3}} ight)$$
$$ \Rightarrow $$   $$	heta $$
2
= $$-$$ $${{\pi  \over 3}}$$
Relative angular displacement of the two particle,
$$	heta $$ = $$	heta $$
1
$$-$$ $$	heta $$
2
= $${{\pi  \over 2}}$$ $$-$$ $$\left( { - {\pi  \over 6}} ight)$$
=  $${{{2\pi } \over 3}}$$
Relative angular velocity $$=$$ $$\omega  - \left( { - \omega } ight)$$ = $$2\omega $$
If they cross each other at time t
then,   t = $${	heta  \over {2\omega }}$$ = $${{2\pi } \over {3 	imes 2\omega }}$$ = $${\pi  \over {3 	imes {{2\pi } \over T}}}$$ = $${T \over 6}$$

Answer

$${T \over 6}$$

Answer

$${5T \over 6}$$

Answer

$${T \over 3}$$

Answer

$${T \over 4}$$

Oscillations

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