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Question:1

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.

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MTH401 Assignment No.2 Solution Spring 2020


Mass= 5kg
Natural length =0.5m
Force = 35.6N
Stretch = 0.5m
X(0) = 0.5
x'(0) = 0 (Initial Velocity)
(Differential equations)
by hook's Law
F=ks
35.6=k(0.5)
k=71.2N/m
Its (Differential equations) Now Find Solution

Complimentary Solution is..

Now Check Initial Conditions..

MTH401-Assignment-2-IDEA-SOL-SPRING-2020

Click on the below Link:

MTH401-Assignment-2-IDEA-SOL-SPRING-2020

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