A 0.600 kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.600 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.5 cm. (Assume the position of the object is at the origin at t = 0.) (a) Calculate the maximum value (magnitude) of its speed and acceleration. _____ cm/s _____ cm/s^2 (b) Calculate the speed and acceleration when the object is 7.50 cm from the equilibrium position. _____ cm/s _____ cm/s^2 (c) Calculate the time interval required for the object to move from x = 0 to x = 5.50 cm. _____ s

Explanation / Answer

w = sqrt(k/m) = sqrt(8/0.6) = 3.65 rad/s

and, we can model the motion as

x(t) = 12. 5cm * sin(3.65*t)

(a) max speed = A*w = 12.5*3.65 cm/s = 45.625 cm/s

max accel = A*w^2 = 12.5*(3.65)^2 = 166.53 cm/s^2

(b) TME = 0.5kA^2 = 0.5* 8.00N/m * (0.125m)^2 = 0.0625 J

When, x = 0.075 m,

PE = 0.5* 8.00N/m * (0.075m)^2 = 0.0225 J

TME - PE = 0.04 = 0.5*0.6*v^2

v = sqrt(0.04/(0.5*0.6)) = 0.3651 m/s = 36.51 cm/s

a = kx / m = 8.00N/m * 0.075m / 0.60kg

a = 1 m/s^2 = 100 cm/s^2

(c) 5.50 cm = 12.5 cm * sin(3.65*t)

3.65*t = arcsin(5.50/12.5) = 0.456

t = 0.456/3.65 = 0.125 s

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