A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is cm, where = .
Answer
12
Solution
vmax = A 10 = 4 = 52 rad/s
v = A2 - x2 5 = 52 16 - x2
16 - x2 = 4 x2 = 12
x = 12 cm, = 12
Answer: 12