A block with a mass of 0.500 kg is connected to a spring, displaced in the posit

Question

A block with a mass of 0.500 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilibrium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 35.0 cm from equilibrium is at t = 0.200 s.

(a) What is the block's period of oscillation?

(b) What is the the value of the spring constant?

(c) What is the block's velocity at t = 0.200 s? (Indicate the direction with the sign of your answer.)

(d) What is the block's acceleration at t = 0.200 s? (Indicate the direction with the sign of your answer.)

Explanation / Answer

equation of oscillation

x = A*cos(wt)

A = 0.5 m

(a)

x = 0.35 at t = 0.2

0.35 = 0.5*cos(w*0.2)

w = 3.98 rad/s

time period = T = 2pi/w = 1.58 s <<--------answer

+++

(b)

K = m*w^2 = 0.5*3.98^2 = 7.92 N/m

+++++

(c)

v = dx/dt = A*w*sinwt

v = 0.5*3.98*sin(3.98*0.2)

v = 1.42 m/s

(d)

a = dv/dt = -A*w^2*cos(wt)

a = -0.5*3.98^2*sin(3.98*0.2)

a = -5.66 m/s^2

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