Question

A mass m=0.65 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has a constant K=85 N/m and negligible mass. At time t=0 the mass is released from rest at a distance d=0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by  y(t) = A cos(ωt – φ). The positive y-axis point upward.

Part (b) Determine the value of the coefficient A, in meters.

Part(d) Enter an expression for the velocity along the y-axis as a function of time, in terms of A, ω, and t using the value for φ from the previous part.

Answer 1

Part b)

A = amplitude = d = distance below equilibrium height = 0.35 m

Pact c)

y(t) = A Cos(wt - )

at t = 0 , y(t) = A

A = A Cos(wt - )

Cos(w(0) - ) = 1

= 0 rad

y(t) = A Cos(wt - )

Taking derivative both side relative to "t"

dy(t) /dt = (d/dt) (A Cos(wt - ))

v(t) = - A w Sin(wt - ))

v(t) = - A w Sinwt