C1-3. SHM Dynamics and Energy

1. Energy Changes in SHM

In an undamped oscillator, energy transfers seamlessly between Kinetic ($E_K$) and Potential ($E_P$) energy. Total Energy ($E_T$) remains perfectly constant.

Energy Equations:HL Only

$$E_K = \dfrac{1}{2} m \omega^2 (x_0^2 - x^2)$$
$$E_T = \dfrac{1}{2} m \omega^2 x_0^2$$

Simple Pendulum: $x, v, a, E_T, E_P$, and $E_K$ vs. $t$ Example

Equilibrium (0) Extreme Left (-x₀) Equilibrium (0) Return (+x₀) a v a v a t = 0 x = +x0 v = 0 a = -amax t = T/4 x = 0 v = -vmax a = 0 t = T/2 x = -x0 v = 0 a = +amax t = 3T/4 x = 0 v = +vmax a = 0 t = T x = +x₀ v = 0 a = -amax Displacement x Velocity v Acceleration a Energy Time t Et 0 T/4 T/2 3T/4 T Total Et Potential Ep Kinetic Ek

Horizontal Spring System: $x, v, a$, and $E_T$ vs. $t$ Example

Equilibrium (0) Extreme Right (+x₀) Equilibrium (0) Extreme Left (-x₀) Equilibrium (0) v a v a v t = 0 x = 0 v = +vmax a = 0 t = T/4 x = +x0 v = 0 a = -amax t = T/2 x = 0 v = -vmax a = 0 t = 3T/4 x = -x0 v = 0 a = +amax t = T x = 0 v = +vmax a = 0 Displacement x Velocity v Acceleration a Energy Time t Et 0 T/4 T/2 3T/4 T Total Et Potential Ep Kinetic Ek

Example 1

How does the period of the kinetic energy oscillation ($T_{KE}$) compare to the period of the oscillating mass ($T$)?

Solution:

  • In a single mechanical cycle, the mass passes the center (maximum speed and max $E_K$) twice—once moving left, once moving right.
  • Therefore, the kinetic energy peaks twice per cycle. The frequency of energy oscillation is double the mechanical frequency ($f_{energy} = 2f$).
  • Conclusion: Since frequency is doubled, the period is halved. $\mathbf{T_{KE} = \dfrac{1}{2}T}$.

Example 2

Problem: A particle oscillates with SHM of period $T$. Which graph shows the variation with time of the kinetic energy of the particle?

A. Ek t T B. Ek t T ✓ C. Ek t T D. Ek t T

Solution:

Graph B is correct. Since kinetic energy must remain strictly positive (it relies on $v^2$) and completes exactly two full cycles within one standard mechanical period ($T$), option B perfectly satisfies the required conditions.