Energy in Oscillating Systems

1. Kinetic Energy in SHM

Kinetic energy (KE) is the energy of motion. For a particle of mass $m$ moving with velocity $v$, the kinetic energy is given by:
$KE=\frac{1}{2}m{v}^2$

We can express this kinetic energy in terms of either displacement ($x$) or time ($t$).

Kinetic Energy as a Function of Displacement

To express KE in terms of displacement $x$, we first need an expression for velocity $v$ in terms of $x$. We can derive this from our standard SHM equations:
From $x=A\cos (\omega t+\varphi )$, we have $\cos (\omega t+\varphi )=\frac{x}{A}$.
From $v=-A\omega \sin (\omega t+\varphi )$, we have $\sin (\omega t+\varphi )=-\frac{v}{A\omega }$.

Using the trigonometric identity ${\sin }^2\theta +{\cos }^2\theta =1$:
${\left(-\frac{v}{A\omega }\right)}^2+{\left(\frac{x}{A}\right)}^2=1$
$\frac{v^2}{A^2{\omega }^2}+\frac{x^2}{A^2}=1$
$v^2={\omega }^2(A^2-x^2)$
So, the magnitude of the velocity is $v=\omega \sqrt{A^2-x^2}$.

Now, substituting this expression for $v^2$ into our KE formula:
$KE=\frac{1}{2}m[{\omega }^2(A^2-x^2)]$
Since $k=m{\omega }^2$, we get the important relation:
$KE(x)=\frac{1}{2}k(A^2-x^2)$

From this equation, we can see:

• Maximum KE: Occurs when $x=0$ (the mean position). $K{E}_{max}=\frac{1}{2}k{A}^2$.
• Minimum KE: Occurs when $x=\pm A$ (the extreme positions). $K{E}_{min}=0$.

The following video provides a clear derivation and explanation of these concepts.

2. Potential Energy in SHM

Potential energy (PE) is the energy stored in a system due to its position or configuration. For a spring-mass system, the restoring force $F=-kx$ is a conservative force, and the associated elastic potential energy is:
$U(x)=\frac{1}{2}k{x}^2$

This is the primary formula for potential energy in SHM. From this, we can see:

• Maximum PE: Occurs at the extreme positions, where $x=\pm A$. $U_{max}=\frac{1}{2}k{A}^2$.
• Minimum PE: Occurs at the mean position, where $x=0$. $U_{min}=0$.

The video below explains this concept.

3. Total Mechanical Energy and Conservation

The total mechanical energy ($E$) of the oscillator is the sum of its kinetic and potential energies:
$E=KE+U$

Let's substitute the expressions we found in terms of displacement $x$:
$E=\frac{1}{2}k(A^2-x^2)+\frac{1}{2}k{x}^2$
$E=\frac{1}{2}k{A}^2-\frac{1}{2}k{x}^2+\frac{1}{2}k{x}^2$
$E=\frac{1}{2}k{A}^2$

This is a powerful result. It shows that the total mechanical energy of a simple harmonic oscillator is constant. It does not depend on the instantaneous position $x$ or time $t$. The energy simply transforms between kinetic and potential forms, but their sum is always conserved.

Notice also that:
$E_{total}=K{E}_{max}\text{ (at }x=0)=U_{max}\text{ (at }x=\pm A)=\frac{1}{2}k{A}^2$
Since $\omega =\sqrt{k/m}$, we can also write the total energy as:
$E=\frac{1}{2}m{\omega }^2{A}^2$

This confirms that the total energy depends only on the mass, angular frequency, and amplitude of the oscillation.

4. Graphical Representation of Energy in SHM

Visualizing how these energies change is essential.

Energy vs. Displacement

The formulas $U(x)=\frac{1}{2}k{x}^2$ and $KE(x)=\frac{1}{2}k(A^2-x^2)$ describe parabolas when plotted against $x$.

From the graph, you can see that at any displacement $x$, the sum of the PE and KE values always equals the total energy $E$.

Energy vs. Time

We can also plot these energies as functions of time, using $x(t)=A\cos (\omega t)$ (assuming $\varphi =0$):

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

To tie all these concepts together, let's turn to your NCERT textbook.

5. Problem-Solving: Applying the Energy Concepts

Now, let's put this theory into practice. Your ability to apply these formulas is key for the NEET exam. We'll work through a standard problem from the NCERT textbook.

Let's break down the logic of Example 13.7:

Given:

• Mass, $m=1$ kg
• Spring constant, $k=50$ N/m
• Amplitude, $A=10$ cm = $0.1$ m
• We need to find energies at displacement $x=5$ cm = $0.05$ m.

Solution Steps:

1. Calculate Total Energy (E): This is the easiest first step and a good check for your final answer. The total energy is constant throughout the motion.
   $E=\frac{1}{2}k{A}^2=\frac{1}{2}(50\text{ N/m})(0.1\text{ m}{)}^2=0.25\text{ J}$

2. Calculate Potential Energy (U) at x = 0.05 m:
   $U=\frac{1}{2}k{x}^2=\frac{1}{2}(50\text{ N/m})(0.05\text{ m}{)}^2=0.0625\text{ J}$

3. Calculate Kinetic Energy (KE) at x = 0.05 m: You could use $KE=\frac{1}{2}k(A^2-x^2)$, or, since you know the total energy, you can simply use the conservation principle:
   $KE=E-U$
   $KE=0.25\text{ J}-0.0625\text{ J}=0.1875\text{ J}$
   (Note: The textbook calculates velocity first and gets $KE\approx 0.19$ J due to rounding. Using the energy conservation formula is often more direct and precise).

This problem demonstrates the power of the energy conservation principle.

For more practice with problem-solving techniques, the following video is an excellent resource.

Conclusion

Today, we've dissected the energy of an object in Simple Harmonic Motion, a cornerstone of your NEET preparation for this chapter.

Key Takeaways:

• Kinetic Energy: $KE=\frac{1}{2}k(A^2-x^2)$. It is maximum at the center ($x=0$) and zero at the extremes ($x=\pm A$).
• Potential Energy: $U=\frac{1}{2}k{x}^2$. It is zero at the center and maximum at the extremes.
• Total Energy: $E=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$. It is always conserved in an ideal system.
• The relationship $E_{total}=KE+U$ is a powerful tool for solving problems, allowing you to find one quantity if you know the other two.

In our next lesson, we will shift our focus from the oscillation of a single particle to see how these oscillations can travel through a medium. We will derive the general equation for a one-dimensional progressive wave, setting the stage for understanding wave phenomena.