What is the equation of motion of a spring?

A 0.70 mass is suspended on a spring that stretches a distance of 12.0 cm. The mass is then pulled down an additional distance of 20 cm and released. What is the equation of motion?



a. y = -20cos(12t)

b. y = -20cos(9t)

d. y = 20cos(9t)

e. none of the above



Please explain how you got your answer. Thank you!What is the equation of motion of a spring?The general form of the eq. is; Where "w" is the frequency of oscilation and "b" is a phase constant determined by the initial conditions. The spring will oscillate about its equilibrium position (where it hangs when at rest) with amplitude "A"

y = ACos(wt + b)



The easy part is that at t=0 you want y= -20cm (taking y=0 at the equilibrium position). Sub t=0 in the general form and y=-20 at that time gives;

-20 = ACos(b)



This will be satisfied if b=0 and A=-20. Put these values back in the general form;

y = -20Cos(wt)



You find "w" from the formula ,w = SqRt(k/m)



You find the spring constant from the equilibrium information;

ky = mg

k/m = g/y = 9.8/.12 = 81.7 1/s^2



Then

w = SqRt(81) = 9.03 rad/s ~ 9 rad/s



And finally

y = -20Cos(9t)

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