Please provide working 2. A 2.50 kg block is attached to an ideal spring, which

Question

Please provide working

2. A 2.50 kg block is attached to an ideal spring, which has a force constant k = 125 Nm^-1. Initially, the block has zero velocity and is displaced 0.15 m from the equilibrium point (xo = 0.15 m). a. Derive an equation for the block's position as a function of time in the form x = Acos (omega t + phi) . (That is, find values for A, omega and phi). (0.15 m, 7.07 rad/s, 0) b. At what time will the block first pass through the equilibrium point (xo = 0 m)? (0.222 s) c. What will the velocity of the block be at the instant it passes through the equilibrium point? (-1.06 m s^-1)

Explanation / Answer

w = sqrt(k/m) = sqrt(125/2.5) = 7.07

since displaced 0.15 from equilibrium with v = 0

A = 0.15 and phi = 0

b)

0 = 0.15*cos(7.07*t)

cos(7.07*t) = 0

first time zero when 7.07*t = pi/2

t = 0.222 s

c) v = dx/dt = - A w sin(wt + phi)

= -0.15*7.07*sin(7.07*0.222) = -1.06

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