Question

An object spring system moving with simple harmonic motion has an amplitude A.

a) What is the total energy of the system in terms of k and A only? (E=1/2kA^2)
b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass m, velocity v, spring constant k, and position x. (1/2mv^2=kx^2)
c) Using the results of parts (a) and (b) and the conservation of energy equation, find the positions x of the object when its kinetic energy equals twice the potential energy stored in the spring.

Answer 1

the total energy is kA^2/2=P.E+k.E=3P.E
This gives P.E=kA^2/6=kx^2/2
Solve it for x

Answer 2

I need part C. I'm not sure how to exactly get it but the answer is: x=A/3^1/2

Answer 3

If you can expand your steps a bit, I would appreciate it.

Answer 4

the total energy is conserved in SHM and is equal to kA^2/2 at all the times.
The total energy is the sum of the potential energy and the kinetic energy.
Hence kA^2/2=P.E+K.E
Given K.E.=2P.E. at some instant and we have to find the value of x at which thia will happen
So kA62/2=P.E+2P.E=3P.E
P.E=kA^2/6
But P.E=kx^2/2
So kx^2/2=kA^2/6
This can be solved for x and x is indeed equal to A/3^1/2

Answer 5

ohhh all right. Thanks! :)