Energy in simple harmonic motion Knowledge-Based Study

Study notes

1. Conservation of Mechanical Energy

   In an ideal simple harmonic oscillator, such as a mass on a frictionless horizontal spring, no non-conservative forces perform work on the system. Consequently, the total mechanical energy of the system remains constant throughout the motion.

   As the object oscillates back and forth about its equilibrium position, energy is continuously transformed between two forms:

   • Kinetic Energy ($K$): Associated with the motion of the oscillating mass.
   • Potential Energy ($U$): Associated with the deformation of the restoring medium (elastic potential energy in a spring).

   At any instant, the total mechanical energy $E$ is given by the sum of kinetic and potential energies:

   $$E = K + U = \text{constant}$$

   Position-Dependent Energy Relations

   Consider a body of mass $m$ attached to a spring of force constant $k$, oscillating with amplitude $A$. When the displacement from equilibrium is $x$, the potential energy stored in the spring is:

   $$U(x) = \frac{1}{2} k x^2$$

   The kinetic energy of the mass with speed $v$ is:

   $$K = \frac{1}{2} m v^2$$

   At the turning points ($x = \pm A$), the instantaneous velocity is zero, meaning all mechanical energy is stored as potential energy:

   $$E = \frac{1}{2} k A^2$$

   Because total mechanical energy is conserved, we can express the kinetic energy directly in terms of displacement:

   $$K(x) = E - U(x) = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)$$

   Equating $K = \frac{1}{2} m v^2$ to $\frac{1}{2} k (A^2 - x^2)$ and noting that $\omega = \sqrt{k/m}$, the speed at any position $x$ is:

   $$v(x) = \omega \sqrt{A^2 - x^2}$$

   Analysis of Critical Points

   The continuous exchange of energy displays distinct values at characteristic points in the oscillation cycle:

   • Equilibrium Position ($x = 0$): Potential energy reaches its minimum ($U = 0$), and kinetic energy reaches its maximum ($K_{\max} = E$). The speed is maximum: $v_{\max} = \omega A$.
   • Extremes or Turning Points ($x = \pm A$): The speed is momentarily zero ($K = 0$), and potential energy is at its maximum ($U_{\max} = \frac{1}{2} k A^2 = E$).
   • Equal Energy Position ($K = U$): When $K = U = \frac{1}{2} E$, we set $\frac{1}{2} k x^2 = \frac{1}{4} k A^2$, which yields: $$x = \pm \frac{A}{\sqrt{2}}$$ At these positions, the total mechanical energy is shared equally between kinetic and potential forms.

   Time Dependence and Oscillation Frequency

   Let the displacement of the oscillator be described by $x(t) = A \cos(\omega t)$, where $\omega$ is the angular frequency. The velocity is then $v(t) = -A \omega \sin(\omega t)$.

   Substituting these into the energy formulas gives:

   $$U(t) = \frac{1}{2} k A^2 \cos^2(\omega t)$$

   $$K(t) = \frac{1}{2} k A^2 \sin^2(\omega t)$$

   Adding both equations confirms conservation of energy via the identity $\cos^2(\theta) + \sin^2(\theta) = 1$:

   $$E = U(t) + K(t) = \frac{1}{2} k A^2$$

   Frequency of Energy Oscillations

   Using trigonometric double-angle identities, $\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}$ and $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$. This shows that both potential and kinetic energy oscillate with an angular frequency of $2\omega$, which is twice the frequency of the displacement and velocity oscillations.

   In one complete cycle of motion (period $T$), kinetic energy and potential energy each reach their maximum values twice.