a 2 kg block attached to a spring undergoes simple harmonic motion described by

Question

a 2 kg block attached to a spring undergoes simple harmonic motion described by x = 30 cm cos [(6.28 rad/s)t + pi/4].

Determine a) amplitude, b) the spring constant, c) the frequency, d) the maximum speed of the block, e) maximum acceleration of the block and f) total energy of the spring block.

I know that at maximum equilibrium t = 0 and phase constant is 0, so amplitude is 30 cm; I'm thinking omega = 6.28 and then omega squared equals spring constant K divided by mass then solve for K. I'm not really sure about the other ones. Thanks for any assistance, I'll rate!

Explanation / Answer

x = 30 cm cos [(6.28 rad/s)t + pi/4].

a) amplitude = 30cm

b) angular frequency = (k/M) = 6.28 rad/s

=> k = 78.88 N/m

c) frequency f = /(2)

=> f = 1 Hz

d) speed ,v = dx/dt = -30 x 6.28 sin((6.28 rad/s)t + pi/4)

max speed = 30*6.28 = 188.4 m/s

e) acceleration a = dv/dt = - 30 * 6.28 * 6.28 cos ((6.28 rad/s)t + pi/4)

max acceleration = 30 * 6.28 * 6.28 = 1183.152 m/s^2

f)total energy = (1/2)mVmax2 = 35494.56 J

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