First slide

Simple hormonic motion

Question

Two particles $P_{1} and P_{2}$ are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time after which the separation between the particles become 6 cm.

Moderate

A

5 sec
B

2 sec
C

4 sec
D

6 sec

Solution

The coordinates of the particles are
$X_{1} = A_{1} \cos ωt, x_{2} = A_{2} \cos ωt$
separation = $x_{1} - x_{2} = (A_{1} - A_{2}) \cos ωt = 12 \cos ωt$
Now $x_{1} - x_{2} = 6 = 12 \cos ωt$
$⇒ ωt = \frac{π}{3} \Rightarrow \frac{2 π}{12} . t = \frac{π}{3}$
$\Rightarrow t = 2 s$

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