Question

A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.6 cm. (Assume the position of the object is at the origin at t0.) (a) Calculate the maximum value of its speed. cm/s (b) Calculate the maximum value of its acceleration cm/s2 (c) Calculate the value of its speed when the object is 9.60 cm from the equilibrium position. (d) Calculate the value of its acceleration when the object is 9.60 cm from the equilibrium position. (e) Calculate the time interval required for the object to move from x = 0 to x 3.60 cm.

Answer 1

ω = √(k/m) = √(8.00N/m / 0.420kg) = 4.36 rad/s

and we can model the motion as

x(t) = 11.6cm * sin(4.36*t)

(a) max speed = A*ω = 50.63 cm/s

(b) max accel = A*ω² = 220.73 cm/s²

(c) TME = ½kA² = ½ * 8.00N/m * (0.116m)² = 0.053824 J

When x = 0.096 m, PE = ½ * 8.00N/m * (0.096m)² = 0.036864 J

leaving KE = 0.01696 J = ½ * 0.420kg * v²

v = 0.284 m/s = 28.4 cm/s

(d) a = kx / m = 8.00N/m * 0.096m / 0.420kg = 1.83 m/s² = 183 cm/s²

(e) 3.6 cm = 11.6cm * sin(4.36*t)

4.36*t = arcsin(3.6/11.6) = 0.316

t = 0.316 / 4.36 = 0.0724 s