The drawing shows a top view of a frictionless horizontal surface, where there a

Question

The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass m1 and m2 attached to them. Each spring has a spring constant of 200 N/m. The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at x = 0 m if (a) m1 = m2 = 5.0 kg and (b) m1 = 5.0 kg and m2 = 45 kg?

Explanation / Answer

the time periods of oscillations of the two bodies are given by t1 = sqrt(m1/k1) and t2 = sqrt(m2/k2) now for case a) when m1= m2 = 5 kg and k1 = k2 = 200 N/m => t1 = t2 = 0.1581 s hence the two masses will be side by side after a time of t = 0.1581 for the first time. b) when m1 = 5 and m2 = 45 => t1 = 0.1581 s and t2 = 0.4743 s => the two masses will be at their initial position at t= 0.4743 s for the first time. NOTE* = the time period is given by t = 2*pi*sqrt(m/k) = 0.993 for the first case hence the two bodies will meet at the mean position at t= T/4 = 0.993/4 = 0.24 s 2) in this case the time period is given by t1 = 0.993 and t2 = 2.98 s now the two bodies will be side by side at t = t2/4 = 0.745 s The answers= .24 and .74

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