A spring with a spring constant of 65 N/m is attached to a 0.3 kg mass. Each cyc

Question

A spring with a spring constant of 65 N/m is attached to a 0.3 kg mass. Each cycle the mass travels a distance of 5.7 cm between its top and bottom extreme positions. Assume the mass starts at its top position.

I have already found the the frequency of vibration: 2.34Hz, the angular frequency of vibration: 14.7196 rad/sec, and the period of vibrations: 0.43 sec. However, I need help finding:

d) What is the amplitude of vibration?

e) Write down the math expressions for position, velocity and acceleration as a function of time. (i.e. s=?, v=?, a=?)

s(t) =

v(t) =

a(t) =

and i) What is the expression for the force acting at any time on the spring?

Explanation / Answer

Here ,

d) amplitude = 5.7/2 = 2.85 cm

e) for the mass

s(t) = A * cos(w * t )

s(t) = 2.85 * cos(14.72 * t) cm

for the velocity

v(t) = - 2.85 * 14.72 * sin(14.72 * t)

v(t) = -41.95 * sin(14.72 * t) cm/s

for the acceleration

a(t) = - 2.85 * 14.72^2 cos(14.72 * t)

a(t) = - 6.17 * cos(14.72 * t) m/s^2

i) for the force at any time

F(t) = m * a (t)

F(t) = 0.3 * (- 6.17 * cos(14.72 * t))

F(t) = -1.85 * cos(14.72 * t) N

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