A 0.540-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.540-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.2 cm. (Assume the position of the object is at the origin at t = 0.) (a) Calculate the maximum value of its speed. 0.302 (b) Calculate the maximum value of its acceleration. cm/s^2 (c) Calculate the value of its speed when the object is 8.20 cm from the equilibrium position. 0.266 (d) Calculate the value of its acceleration when the object is 8.20 cm from the equilibrium position. cm/s^2 (e) Calculate the time interval required for the object to move from x = 0 to x = 4.20 cm. s

Explanation / Answer

(a) Maximum speed, vmax = A = A(k/m)1/2 = 10.2 * (8.00 / 0.540)1/2 = 39.3 cm/s

(b) Maximum acceleration, amax = A2 = A(k/m) = 10.2 * (8.00 / 0.540) = 151 cm/s2

(c) Equation of motion is of the form of: x = Asin(t + )

Since the object is at origin at t = 0, we have,

x = Asint,

where = (8.00 / 0.540)1/2 = 3.849 rad/s, A = 10.2 cm.

At x = 8.20 cm, sint = x/A = 8.20 / 10.2 = 0.804

Now, v = dx/dt = A cost = vmax cost

So, at x = 8.20 cm, v = 39.3 * [1 - 0.8042]1/2 = 23.4 cm/s

(d) Acceleartion, a = dv/dt = -A2 sint

At x = 8.20 cm, a = -151 * 0.804 = -121 cm/s2

(e) We have, x = A sint

So, time taken to travel to 4.20 cm is,

t = (1/) sin-1(x/A) = (1 / 3.849) * sin-1(4.20 / 10.2) = 0.110 s

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