Asked by Lisa
Now suppose that the weight attached to the spring is pulled down a further 5cm from its equilibrium positions and released from rest. Show that newtons second law leads to the dynamic equation (in SI units)
d^2z/dt^2 + 36z - g = 0
where z(t) measures the extension of the spring from its natural lenght and g is the acceleration due to gravity.
The first part of this question was related to Hooke's Law which i have answered correctly, i am just unsure how you start this particular question.
d^2z/dt^2 + 36z - g = 0
where z(t) measures the extension of the spring from its natural lenght and g is the acceleration due to gravity.
The first part of this question was related to Hooke's Law which i have answered correctly, i am just unsure how you start this particular question.
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