A 10 Kilogram object suspended from the end of a vertically hanging spring stret
ID: 3160786 • Letter: A
Question
A 10 Kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t = 0, the resulting mass-spring system is disturbed from its rest state by the force F(t) = 80 cos (10t) The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds. Determine the spring constant k. k = Newtons/meter Formulate the initial value problem for y(t), where y(t) is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (Give your answer in terms of y, y', y", t. Differential equation: help (equations) Initial conditions: y(0) = and y'(0) = help (numbers) Solve the initial value problem for y(t) y(t) = help (formulas) Plot the solution and determine the maximum excursion from equilibrium made by the object on the time interval 0 lessthanorequlato tExplanation / Answer
a.To determine the Spring constant ...Hooks law states that force needed to extended or compress a spring by some distance is directly proprtional to the distance.
ie F =KX
Where F is force applied in N/m2.
K is the spring constant
x is the distance measured in meter.
to calculate the spring constant K
using the above expression
K=F/x
F(t)=80cos(t)
t=0
F(0)=80cos(0)
F=80N/m2.
x=9.8cm=9.8*10-2m.
Spring constant is given by K=F/x
K=80/9.8*10-2
K=816N/m
spring constant is found to be 818N/m.
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