Question

2 particles undergoing SHM start from the mean position and go in opposite directions. Particle 1 starts with a speed of 10m/s and particle 2 starts with a speed 0f 5m/s. If the amplitude(=10cm) is the same. At what position will they first meet?

(a) 0.0866m

(b) -0.0633m

(c) 0

(d) 0.0633m

I got this question in an international level competition.

I would like to ask this question from Simple Harmonic Motion in portion Oscillations of Physics – Class 11

Answer 1

Correct option is (a) 0.0866m

The explanation: Let the first particle go towards negative displacement.                                                                                       Its equation will be:                                                                                                                                                                                x1 = 0.1sin(w1t + π) & maximum speed = 0.1w1=10

W1 = 100s^-1.

Let the second particle go towards positive displacement.                                                                                              Its equation will be:                                                                                                                                                               x2 = 0.1sin(w2t) & maximum speed = 0.1w2 = 5

W2 = 50s^-1.

Now, in their phasor diagram: w1t + w2t =2π.

∴ t = 2π/150.

At t = π/150

x1 = 0.1sin(50*2π/150) = 0.0866m.