When the displacement is one-half the amplitude in SHM The fraction of the total energy to that of potential energy is?

The total energy is given by $E = \frac{1}{2} k x_{m 2}$ , where $k$ is the spring constant and $x_{m}$ is the amplitude. We use the answer from part (b) to do part (a), so it is best to look at the solution for part (b) first,

(a) The fraction of the energy that is kinetic is

$\frac{K}{E} = \frac{E - U}{E} = 1 - \frac{U}{E} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$ .

where the result from part (b) has been used.

(b) When $x = \frac{1}{2} x_{m}$ the potential energy is $U = \frac{1}{2} k x^{2} = \frac{1}{8} k m_{m 2}$ . The ratio is

$\frac{U}{E} = \frac{k x _{m 2} /8}{k x _{m 2} /2} = \frac{1}{4} = 0.25$ .

(c) Since $E = \frac{1}{2} k x_{m 2}$ and $U = \frac{1}{2} k x^{2}$ , $U / E = x^{2} / x_{m 2}$ . we solve $x^{2} / x_{m 2}$ for $x$ . We should get $x = x_{m} / 2$ .

The figure below depicts the potential energy (solid line) and kinetic energy (dashed line) as a function of time, assuming $x (0) = x_{m}$ . The two curves intersect when $K = U = E /2$ , or equivalently, $cos^{2} ω t = sin^{2} ω t = 1/2$ .