Physics harmonic motion

Question

 A 150-g object is fixed to the end of a spring that has a spring constant of 15.0 N/m. The object is displaced 15.0 cm to the right and released from rest at t = 0 to slide on a horizontal, frictionless table

 1) Calculate the first time when the object is at the equilibrium position.

2)What is the first time when the object is 10.0 cm to the left of equilibrium?

 3)What is the first time that the object is 5.00 cm to the right of equilibrium, moving toward the left?

Answer 1

With all these "what time is it when x = " questions, it seems like writing the equation of motion would be good. The general equation is x(t) = Acos(ωt) where ω = √(k/m) and A is the Amplitude (amount you "displace it")

Make sure to switch the units to standard SI units so m = 0.15kg and x = 0,15m

For this sketch:

we can write x(t) = 0.15cos(√(15/.15)t) to model the motion.

a) So, to find when x = 0 (the equilibrium point we make x(t) = 0 and solve for t

0 = 0.15cos(√(15/.15)t)

cosine is zero when √(15/.15)t = π/2

10t = π/2

t = π/20 = 0.157 s

b) 10 cm left means x(t) = -.1

-.1 = 0.15cos(√(15/.15)t)

-.1/.15 = cos(√(15/.15)t)

-0.6667 = cos(√(15/.15)t)

cos^-1(-0.6667) = √(15/.15)t

t = cos^-1(-0.6667)/√(15/.15) = 0.230 s

c) 5 cm right of equilibrium moving left is when x = .05 (first portion of the first cycle)

0.05 = 0.15cos(√(15/.15)t)

0.05/.15 = cos(√(15/.15)t)

0.3333 = cos(√(15/.15)t)

cos^-1(0.3333) = √(15/.15)t

t = cos^-1(0.3333)/√(15/.15) = 0.123 s