A mass of 0.22 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.22 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) = (0.30 m)cos[(14 rad/s)t].

Determine the following.

(a) amplitude of oscillation for the oscillating mass
m

(b) force constant for the spring
N/m

(c) position of the mass after it has been oscillating for one half a period
m

(d) position of the mass one-third of a period after it has been released
m

(e) time it takes the mass to get to the position x = ?0.10 m after it has been released
s

Explanation / Answer

x = 0.260 m

m = 7.55 kg

3 *kx = mg

k = mg/3x

= 7.55*9.8/(3*0.260

= 94.86 N/m

Time period = 2*pi*sqrt(k/m)

   = 22.27 sec

C. 1/2 *k*x^2 = 1/2 *m*v^2

   94.86 *0.260^2 = 7.55 *v^2

v = 0.92 m/s

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