a particle executes simple harmonic motion with a time period of 16s. at time t= 2s the particle crosses the mean position while at t= 4s its velocity us 4m/s. what is the amplitude of the motion in meter
Answers
The amplitude of the motion in meter is (32√2)/П m.
We know, according to simple harmonic motion,
y = a sin (ωt + Ф)
Ф is the phase shift
y is the displacement and a is the amplitude.
Again, the angular frequency ω is equal to 2П/T,where T is the time period.
So, y = a sin(2П/T)t
Given time period T = 16s.
ω = 2П/16 = П/8
Also, at t=2s, particle crosses the mean position. It means at t=2s, y = 0.
0 = a sin (2П/8 + Ф)
⇒ sin (П/4 + Ф) = sin П
Comparing both sides, we get,
(П/4 + Ф) = П
⇒ Ф = П - П/4 = 3П/4
So, phase shift is 3П/4.
Again, we know velocity v = dy/dt
So, differentiating y with respect to time, we get:
dy/dt = a cos (Пt/8 + Ф) × (П/8)
v = (aП/8) cos (Пt/8 + Ф)
Putting v = 4 and Ф = 3П/4, and at t = 4s, we get:
4 = (aП/8) cos (4П/8 + 3П/4)
⇒ 4 = (aП/8) cos (5П/4)
⇒ (aП/8) × (-1/√2) = 4 [cos (5П/4) = -1/√2]
⇒ aП/(8√2) = -4
⇒ a = (-32√2)/П
Taking the modulus of amplitude 'a', we get
|a| = (32√2)/П m