A mass of 0.36 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.36 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.36 m)cos[(16 rad/s)t]. Determine the following. amplitude of oscillation for the oscillating mass m force constant for the spring N/m position of the mass after it has been oscillating for one half a period m position of the mass one-third of a period after it has been released m time it takes the mass to get to the position x = -0.10 m after it has been released s

Explanation / Answer

(a) Amplitude of oscillation = 0.36 m

(b) = (k/m)1/2 = 16

=> force constant of the spring, k = 162 * 0.36 = 92.16 N/m

(c) Time period of oscillation, T = 2/ = 2/16 = /8 s

At t = T/2 = /16 s,

x = 0.36 * cos(16 * /16) = 0

So, the particle is at mean position.

(d) At t = T/3 = /24 s,

x = 0.36 * cos(16 * /24) = -0.18 m

(e) -0.10 = 0.36 * cos16t

=> t = (1/16) * cos-1(-0.1/0.36) = 0.116 s

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