Engineering
Physics
Basics of Simple Harmonic Motion
Question

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then

amplitude of motion is 4a

amplitude of motion is 3a

time period of oscillations is 8τ

time period of oscillations is 6τ

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Solution

R – cos θ = a

R – R cos 2θ = 3a

1cosθ1cos2θ=13

1cosθ2sin2θ=13

3 – 3 cosθ = 2 – 2 cos2θ

2 cos2θ – 3 cosθ + 1 = 0 θ = 60º

2cos2θ – 2cosθ – cosθ + 1 = 0

2cosθ (cosθ – 1) – 1

θ = 60º