The spring shown in the image below has an unstretched length of 3.00m, and a spring constant of 75.0N/m. The mass of m=7.50kg is attached to the end of the spring, and the mass is pulled down to stretch the spring to 4.50m in length. This began the mass moving up and down with simple harmonic motion (SHM). In this simple harmonic motion;
a) What is the spring length;
i) At the locations where the velocity of the mass is zero (2)
ii) At the location where its speed is maximized (1)
iii) when its elastic energy is “0J” (1)
b) Determine the acceleration of the mass when the spring is 4.20m long.
c) Determine the speed of the mass when the spring length is 4.30m long.
d) Determine the time it will take the mass to bounce up to the maximum
height and back to its lowest height eight times.
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1 answer

F = k x
omega = w = 2 pi f = 2 pi/T = sqrt (k/m) = sqrt 10
say g = 10 m/s^2
then for x of center of sinusoidal motion = m g/k = 7.5 * 10 / 75 = 1 meter of stretch due purely to gravity
so
answer to first part of first question is 4 meters plus or minus 0.5meters
answer to second part of first question is 3 + 1 = 4 meters spring length of center of sinusoidal motion where speed is max
then
say x = 4.0 + 0.5 sin w t = 4.0 + 0.5 sin [ (sqrt 10 )* t]
then v = 0.5 sqrt 10 * cos [ (sqrt 10 )* t]
and a =  5 sin [ (sqrt 10 )* t]
f = omega/2 pi = sqrt 10/ 2pi
so
T = 1/f = 2 pi / sqrt 10 👍
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answered by Anonymous
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