A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 6.10 N is applied. A 1.600 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is pulled horizontally so that it stretches the spring 5.00 cm and is then released from rest at t = 0.
(a) What is the force constant of the spring?


N/m

(b) What are the angular frequency (?), the frequency, and the period of the motion?
?

rad/s
frequency

Hz
period

s

(c) What is the total energy of the system?


J
(d) What is the amplitude of the motion?


cm

(e) What are the maximum velocity and the maximum acceleration of the particle?
maximum velocity

m/s
maximum acceleration

m/s

(f) Determine the displacement, x, of the particle from the equilibrium position at t = 0.500 s.


cm

Explanation / Answer

a) hooke's law k = 6.10 N / .03 m = 203 N/m b) angular frequency is root (k/m), frequency is angular frequency / 2pi, period is 1/frequency... c) U = 0.5 k x^2. at time 0, all you have is spring potential energy. so total energy = 0.5 * 203 N/m * .05m ^ 2 = 0.0254 J since you stretched it 5cm, that's as far as it goes, amplitude .05 m e) you have max acceleration right away, a=k * x / m, 203*.05/1.100 max speed, is if you convert all PE to KE, 0.0254 J = 0.5 * 1.100 * v^2, solve for V f) you know the amplitude, and the angular frequency, plug this into the equation of motion for springs. x = .05 cos (AngularFrequency*t) and now plug in t=0.5 s

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