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The equation of motion of a particle started at t = 0 is given by x = 5 sin (20 t + 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?

The equation of motion of a particle started at t = 0 is given by x = 5 sin (20 t + 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?

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Sol. . 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.
ankit singh
askIITians Faculty 614 Points
3 years ago

X=5sin(20t+3π​)
v=dtdx​=100cos(20t+3π​)
a=dtdv​=−2000sin(20t+3π​)
v=0 for the first time when,
20t+3π​=2π​
∴   t=120π​ s

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