Q.Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.

The key idea is the Classical Turning Points, which are the extreme positions where the kinetic energy of the oscillator becomes zero, and its potential energy is maximum.

1. The potential energy (PE) of a simple harmonic oscillator is given by $U = \frac{1}{2} k x^2$, where $k$ is the spring constant and $x$ is the displacement from equilibrium.
2. The kinetic energy (KE) is $K = \frac{1}{2} m v^2$. Using the relation $v = \omega \sqrt{A^2 - x^2}$ and $k = m\omega^2$, this becomes $K = \frac{1}{2} k (A^2 - x^2)$, where $A$ is the amplitude.
3. The total mechanical energy (TE) is the sum of PE and KE: $E = U + K = \frac{1}{2} k x^2 + \frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k A^2$.
4. This shows that PE varies parabolically with $x$ (minimum at $x=0$, maximum at $x=\pm A$), KE varies as an inverted parabola (maximum at $x=0$, zero at $x=\pm A$), and TE remains constant, independent of displacement. …

When an object undergoes Simple Harmonic Motion (SHM), its energy continuously transforms between kinetic and potential forms. Understanding this energy variation with displacement is crucial for grasping the dynamics of an oscillator. The key idea is that while kinetic energy (KE) and potential energy (PE) change, their sum, the total mechanical energy (TE), remains constant, assuming no damping forces.

The restoring force in SHM always acts towards the equilibrium position. This force is conservative, meaning that the work done by it is stored as potential energy. As the oscillator moves, its speed and position change, leading to a continuous exchange between KE and PE.

Let's derive the expressions for these energies and then describe their graphical representation.

Deriving the Energy Expressions

1. Displacement and Velocity in SHM For a simple harmonic oscillator, the displacement $x$ from the equilibrium position at any time $t$ can be described by:

where $A$ is the amplitude (maximum displacement) and $\omega$ is the angular frequency.

The velocity $v$ of the oscillator is the time derivative of its displacement:

2. Kinetic Energy (KE) The kinetic energy of the oscillator is given by the standard formula:

Substituting the expression for $v(t)$:

To express KE in terms of displacement $x$, we use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$.
Since $x = A \cos(\omega t)$, we have $\cos(\omega t) = \frac{x}{A}$, and thus $\cos^2(\omega t) = \frac{x^2}{A^2}$.
So, $\sin^2(\omega t) = 1 - \frac{x^2}{A^2}$.
Substituting this into the KE expression:

This equation shows that KE is maximum at the equilibrium position ($x=0$), where $KE_{max} = \frac{1}{2}m\omega^2 A^2$, and zero at the extreme positions ($x=\pm A$).

3. Potential Energy (PE)

The restoring force in SHM is given by Hooke's Law: $F = -kx$, where $k$ is the spring constant.

The potential energy stored in the system is the work done against this restoring force to displace the object by $x$.

For SHM, the angular frequency $\omega$ is related to the mass $m$ and spring constant $k$ by $\omega = \sqrt{\frac{k}{m}}$, which means $k = m\omega^2$.
Substituting $k = m\omega^2$ into the PE expression:

This equation shows that PE is zero at the equilibrium position ($x=0$) and maximum at the extreme positions ($x=\pm A$), where $PE_{max} = \frac{1}{2}m\omega^2 A^2$.

> [!WARNING]
> A common mistake is to assume PE is always $\frac{1}{2}kx^2$. While true for a spring, for a general SHO, it's more accurate to use $PE = \frac{1}{2}m\omega^2 x^2$, as it directly relates to the fundamental parameters of SHM.

4. Total Energy (TE)

The total mechanical energy of the oscillator is the sum of its kinetic and potential energies:

Substituting the expressions for KE and PE in terms of $x$:

Since $m$, $\omega$, and $A$ are constants for a given oscillator, the total energy $TE$ is constant and independent of displacement $x$. This demonstrates the conservation of mechanical energy in an ideal simple harmonic oscillator.

> [!FORMULA]
> The total energy of a simple harmonic oscillator is given by:
> $$TE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$$

The Energy-Displacement Graph

Now, let's describe the graph showing the variation of PE, KE, and TE with displacement $x$. The displacement $x$ ranges from $-A$ to $+A$.

1. Potential Energy (PE) Curve:

   • The equation $PE = \frac{1}{2}m\omega^2 x^2$ is a parabolic function of $x$.
   • It is symmetric about the $y$-axis (PE axis).
   • PE is minimum (zero) at $x=0$ (equilibrium position).
   • PE is maximum at $x=\pm A$ (extreme positions), where $PE_{max} = \frac{1}{2}m\omega^2 A^2$.
   • The curve opens upwards.

2. Kinetic Energy (KE) Curve:

   • The equation $KE = \frac{1}{2}m\omega^2 (A^2 - x^2)$ is also a parabolic function of $x$.
   • It is symmetric about the $y$-axis. …

Step 1: Take $x=A\cos\omega t$, so $v=-A\omega\sin\omega t$.

Step 2: PE $=\tfrac12kx^2=\tfrac12m\omega^2x^2$ — an upward parabola in $x$, zero at $x=0$, maximum $\left(=\tfrac12m\omega^2A^2\right)$ at $x=\pm A$.

Step 3: KE $=\tfrac12mv^2=\tfrac12m\omega^2(A^2-x^2)$ — a downward (inverted) parabola, maximum $\left(=\tfrac12m\omega^2A^2\right)$ at $x=0$, zero at $x=\pm A$. …