Simple Harmonic Motion ( Vertical Spring)

A spring with spring constant 150 N/m hangs from a ceiling. A 2.00 kg mass is attached to the end of the spring and released.
a) How far below its unstretched position will the mass be in equilibrium?
b) How far below its unstretched position will the mass fall?

The answer to part A = 0.1308m but I dont know what to do for part b. I think it might be twice the length of the previous answer but I dont know how to prove that, any help would be greatly appreciated.

a) kx =mg
150 * x = 2 * 9.81
x = 0.1308m
answer
b) distance below its unstretched position will the mass fall = amplitude of SHM
spring force at amplitude = k ( a+ 0.1308) = 150 ( a + 0.1308)
= m * acc
= 2 * acc
acc = 75 (a + 0.1308)
u =0 ; distance = a
v^2 = u^2 + 2 *acc * distance
v^2 = 0 + 2 * 75 (a + 0.1308) * a ---(1)

angular frq w = sq root [ k/m] = sq root [150 / 2] = 8.66
v = w *a
= 8.66 a
sub into (1)
( 8.66 a)^2 = 150 (a + 0.1308) * a
75 a^2 - 150 (a + 0.1308) * a =0
a [ 75 a - 150a - 19.62] =0
a=0 not valid
-75 a - 19.62 =0
a = - 0.26m
the mass will fall 0.26m below its unstretched position
answer
Note: negative sign denotes downward distance.

the position of the mass as a function of time
y = a cos (wt + C)
y = 0.26 cos (8.66 t + C)
determine C when y=0, t=0
hope that will help :)

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