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) 0.0633mI 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

2 particles undergoing SHM start from the mean position and go in oppo

1.	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) 0.0633mI 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» 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.

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