A small block (mass 250 g) hangs from a spring (spring constant =10 N/m). At t =

Question

A small block (mass 250 g) hangs from a spring (spring constant =10 N/m). At t = 0 s the
block is 10 cm above the equilibrium position and is moving down at 1.0 m/s.
• Draw a picture and be sure to indicate a coordinate system and the positive direction.
• Write the position as a function of time, x(t). (This should be of the form x(t) = … and have
numbers except for t on the right-hand side.)
• What are the position and velocity at t=1.2 s?
• What is the block's velocity when it is going through equilibrium on the way up?

Explanation / Answer

m = 250 g, k = 10 N/m, angular frequency w = sqrt(k/m) = sqrt(40) = 6.324 s^-1 At t = 0 s, x(0) = +10 cm, v(0) = -1.0 m/s x(t) = Acos(wt + b) x(0) = Acosb 0.1 = Acosb (1) x'(t) = v(t) = -wAsin(wt + b) v(0) = -wAsinb -1.0 = -wAsinb 1.0/w = Asinb (2) from (1), (2) A = sqrt[0.1^2 + (1.0/w)^2] = 0.187 m b = arctan(1.0/0.1w) = 1 rad so x(t) = 0.187 cos(6.324t + 1) at t=1.2 s x(1.2) = -0.125 m v(1.2) = -0.187*6.324 sin(6.324*1.2 + 1) = -0.877 m/s when it is going through equilibrium on the way up, x(t) = 0, v(t) > 0 cos(6.324t + 1) = 0, sin(6.324t + 1) = -1 v(t) = 0.187*6.324 = 1.18 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.