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.
Please Discuss here about this assignment.Thanks
Our main purpose here discussion not just Solution
Students having same subject can start discussion here to solve assignment, GDB & Quiz and can clear their concepts until solution is provided.
P.S: Please always try to add the discussion in proper format title like “CS101 Assignment / GDB No 01 Solution & Discussion Due Date: ___________”
Then copy Questions from assignment file and paste in Discussion.
+ http://bit.ly/vucodes (For Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (For Past Papers, Solved MCQs, Short Notes & More)
+ Click Here to Search (Looking For something at vustudents.ning.com?)
+ Click Here to Join (Our facebook study Group)
MTH401 Assignment No.2 Solution Spring 2020
Natural length =0.5m
Force = 35.6N
Stretch = 0.5m
X(0) = 0.5
x'(0) = 0 (Initial Velocity)
by hook's Law
Its (Differential equations) Now Find Solution
Complimentary Solution is..
Now Check Initial Conditions..
Click on the below Link: