Dynamic fluids question

Question

The mass in a mass-spring system is pulled 0.03102 m to the right of equilibrium and released. If the spring constant is 36.3 N/m and the mass is 0.402 kg, write a formula for the acceleration of the mass as a function of time. Enter both the amplitude of the acceleration and the angular frequency to the nearest tenth

Answer 1

Let

A = 0.03102 m be maximum displacement,

m = 0.402 kg be the mass,

k = 36.3 N/m be the spring constant,

w = sqrt(k/m) be the angular frequency,

x(t) be the displacement as a function of time.

I assume that you know that

x(t) = A cos(wt)

If you do not, please let me know and I will

show you how to derive it.

Acceleration is then the second derivative of x.

x'(t) = -A w sin(wt)

x"(t) = -A w² cos(wt)

Plugging in the numbers:

w = sqrt((36.3 N/m) / (0.402 kg)) = 9.5 s^-1

x"(t) = -0.03102 . 90.3 cos(9.5t) = -2.8sin(9.5t) m/s^-2