Question

A 180 g mass attached to a horizontal spring oscillates at a frequency of 2.80 Hz. At t = 0 s, the mass is at x=7.00 cm and has vx= -29.0 cm/s. Determine: The maximum speed, the maximum acceleration, the total energy, and the position at t = 4.40 s.

Answer 1

Given, mass m = 180 g = 0.180 kg, frequency f = 2.80 Hz, At t = 0 s, the displacement of the mass is x0 = 7.00 cm = 0.07 m and speed at t = 0s is vx0 = -29.0 m/s.

Angular frequency = 2f = 2(2.80 Hz) = 17.58 rad/s

Spring constant k = m2 = (0.180 kg)(17.58 rad/s)2 = 55.63 N/m

Using energy conservation, total mechanical energy is 1/2 kA2 = 1/2kx02 + 1/2mvx02

A2 = x02 + (m/k).vx02 = (0.07 m)2 + (0.180kg) (-29.0 m/s)2/(55.63 N/m) = 2.726 m2

or A = 1.651 m

The expression for the displacment of the mass is x(t) = A cos(t + ) -------------- (1)

Where is the phase constant, which is comes from A cos = x0

Then, = cos-1(x0/A) = cos-1(0.07/1.651) = 1.52 rad.

At t = 4.40 s, x(t) = (1.651 m) cos[(17.58 rad/s)(4.40 s) + 1.52 rad] = -1.56 m

Velocity at time t is v(t) = -A sin(t +) ------------- (2)

Then maximum velocity is vmax = A -------- (2a)

Therefore, vmax = (17.58 rad/s) (1.651 m) = 29.02 m/s.

Accelerattion at time t is a(t) = -2A cos(t + ) ------------ (3)

Then maximum acceleration is amax = 2A ---------- (3a)

Therefore, amax = (17.58 rad/s)2 (1.651 m) = 510.25 m/s2.