The equation of motion of a particle started at t=0 is given by x=5.sin(20t+π/3) where x is in centimeter and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed
Answers
Given equation,
x = 5Sin(20t + π/3)
(a). When the Particle comes to rest, then, velocity becomes zero.
∴ v = 5 × 20 Cos(20t + π/3)
0 = Cos(20t + π/3)
Cos(20t + π/3) = Cos90
On comparing,
20t + π/3 = 90
20t + π/3 = π/2
20t = π/2 - π/3
20t = π/6
t = π/120 seconds.
(b). a = 5 × 400 × -Sin(20t + π/3)
Now, a = 0,
Sin(20t + π/3) = Sin0 or Sinπ
Therefore, t = -π/60 or π/30 s.
Therefore, t = π/30 seconds.
(c). In S.H.M., Velocity will be maximum where acceleration is zero. therefore, t = π/30 seconds.
Hope it helps.
x = 5 sin (20t + π/3)
a) Max. displacement from the mean position = Amplitude of the particle. At the extreme position, the velocity becomes ‘0’.
∴ x = 5 = Amplitude.
∴ 5 = 5 sin (20t + π/3) sin (20t + π/3) = 1 = sin (π/2)
⇒ 20t + π/3 = π/2
⇒ t = π/120 sec.,
So at π/120 sec it first comes to rest.
b) a = ω2x = ω2 [5 sin (20t + π/3)] For a = 0, 5 sin (20t + π/3) = 0
⇒ sin (20t + π/3) = sin (π)
⇒ 20 t = π – π/3 = 2π/3
⇒ t = π/30 sec.
c) v = A ω cos (ωt + π /3) = 20 × 5 cos (20t + π/3)
when, v is maximum i.e.
cos (20t + π/3) = –1 = cos π
⇒ 20t = π – π/3 = 2 π/3
⇒ t = π/30 sec.