A) [tex]x=\pm \frac{A}{2\sqrt{2}}[/tex]
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
[tex]E=\frac{1}{2}kA^2[/tex] (1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex] (2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
[tex]U=\frac{1}{3}K[/tex]
Using (2) we can rewrite this as
[tex]U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}[/tex]
And using (1), we find
[tex]U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2[/tex]
Substituting [tex]U=\frac{1}{2}kx^2[/tex] into the last equation, we find the value of x:
[tex]\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}[/tex]
B) [tex]x=\pm \frac{3}{\sqrt{10}}A[/tex]
In this case, the kinetic energy is 1/10 of the total energy:
[tex]K=\frac{1}{10}E[/tex]
Since we have
[tex]K=E-U[/tex]
we can write
[tex]E-U=\frac{1}{10}E\\U=\frac{9}{10}E[/tex]
And so we find:
[tex]\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A[/tex]
