# Thread: Simple Harmonic Motion 2

1. ## Simple Harmonic Motion 2

A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.

2. Originally Posted by UltraGirl
A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.
ignore this , see next .

3. Originally Posted by UltraGirl
A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.
Since the particle in SHM passes through two points A and B, 22 cm apart, with the same speed, they must be in the opposite sides of the equilibrium. It takes one second to reach B from the equilibrium position. The particle moves further, reaches the maximum amplitude and returns back to B in 2 seconds. So the particle takes 2 seconds to reach the maximum position from the equilibrium position.
Hence the period of SHM is 8 seconds.
Equation of SHM is
A = Aosin(ωt)
In one second the amplitude is 11 cm. So
11 = Aosin(2π/8)
Ao = 11*sqrt(2)