Highest Point Reached By a Spring??

Question

A mass of 0.500 kg is hung from a spring and allowed to come to rest. The mass is then pulled down 4 cm from this rest position and released from rest at t=0.; the mass then oscillates with a period of 0.2 s. What is the y value for the highest point reached by the mass and the time it took to reach it?? What would the equation of motion for the mass be in terms of y(t), substituting the values for the constants and specifying the units? What is the spring constant in this case?

Answer 1

y max = 4 cm

m = 0.5 kg

T = 0.2 s

time to reach y max  is 1/4 period = (1/4)(0.2s) = 1/8 s

angular frequency =w= (2pi)f = 10pi = 31.4 rad/s

y(t) = 0.04sin(31.4t)

w = sqrt(k/m)

therefore

k =mw^2= 0.5(31.4)^2

k =492.98 N/meter

Answer 2

If y = 0 is the equilibrium point, then clearly y = +4 cm = ymax

Since the mass goes through ½ a cycle from y = -4 mc to y = +4 cm, and a whole cycle takes .2 sec, obviously the time from ymin to ymax is .10 sec

k = 4π²m/T² = 493.48 N/m

w = √(k/m) = 31.42 rad/sec

Eq:  y(t) = 4*sin(w(t -.05)) cm .....You have to subtract .05 sec because the motion started 1/4 cycle (.05 sec) before y = 0.

All these equations are based on Newton's laws.