a) Potential energy: [tex]\frac{1}{9}E[/tex]
b) Kinetic energy: [tex]\frac{8}{9}E[/tex]
c) [tex]x=\pm \sqrt{\frac{2}{3}}A[/tex]
Explanation:
a)
The total energy of a simple harmonic oscillator is given by:
[tex]E=\frac{1}{2}kA^2[/tex]
where
k is the spring constant
A is the amplitude (the maximum displacement) of the system
The potential energy can be written as
[tex]U=\frac{1}{2}kx^2[/tex]
where
x is the displacement
Here, the position is 1/3 of the amplitude, so
[tex]x=\frac{1}{3}A[/tex]
This means that the potential energy is:
[tex]U=\frac{1}{2}k(\frac{1}{3}A)^2=\frac{1}{9}(\frac{1}{2}kA^2)=\frac{1}{9}E[/tex]
b)
The total energy is conserved during the motion, and therefore, at any instant, it is sum of the kinetic energy and the potential energy of the system:
[tex]E=K+U[/tex] (1)
where
K is the kinetic energy
U is the potential energy
As we found in part a), the potential energy when the position is 1/3 of the amplitude is
[tex]U=\frac{1}{9}E[/tex] (2)
Therefore, substituting (2) into (1) and solving for K, we find the kinetic energy of the system:
[tex]K=E-U=E-\frac{1}{9}E=\frac{8}{9}E[/tex]
c)
Here we want to find the value of the position at which the kinetic energy equal one half the potential energy.
Mathematically, this means that
[tex]K=\frac{1}{2}U[/tex]
Using again the formula of the total mechanical energy
[tex]E=K+U[/tex]
And substituting K, we find
[tex]E=\frac{1}{2}U+U = \frac{3}{2}U\\U=\frac{2}{3}E[/tex] (2)
This means that the kinetic energy is one half the potential energy when the potential energy is 2/3 of the total energy.
Using the expression for the total energy:
[tex]E=\frac{1}{2}kA^2[/tex]
And the expression for the potential energy:
[tex]U=\frac{1}{2}kx^2[/tex]
And combinng them into (2), we find:
[tex]\frac{1}{2}kx^2 = \frac{2}{3} (\frac{1}{2}kA^2)\\x^2 = \frac{2}{3}A^2\\x=\pm \sqrt{\frac{2}{3}}A[/tex]
