Question

An obnoxious acquaintance says that at the positions of x and –x
the kinetic energy of a frictionless spring/mass system is the same. This
acquaintance becomes even more obnoxious than usual by saying that the
same is true of the potential energy and also of the total system energy:

a) the acquaintance is correct on all three counts
b) the acquaintance is wrong because total system energy changes
c) the acquaintance is wrong because potential energy is different at
x than it is at –x
d) the acquaintance is wrong because kinetic energy is different at

Answer 1

If we're defining 0 to be at the relaxed length of the spring, then they'd be correct on all accounts.

We know that energy is conserved, so we can look at just the spring potential energy to see how it changes.
Using Hooke's Law:
\[F = kx\]
the definition of work:
\[W = \int\limits_a^bF d^3r\]
and the definitions of potential energy (spring potential energy, in this case), and total energy:
\[V_{spring} = -W_{spring}\]\[E = T + V\](With T being kinetic energy).
We immediately see that the potential energy
\[\int\limits_0^x kydy = \frac{1}{2}kx^2\]
and
\[\int\limits_0^{-x}kydy = \frac{1}{2}k(-x)^2 = \frac{1}{2}kx^2\]
So, the potential energy is the same at both x and -x. Since energy is conserved, this means that so is the kinetic energy.

Answer 2

thanks so much, you explanation helped a lot!