For a body executing SHM
A. potential energy is always equal to its kinetic energy.
B. average poten

Q: Why isn't the average energy equal over *any* time interval?

Because the energy distribution is not uniform. Near the mean position, the particle spends less time but has high kinetic energy. Near the extreme positions, it spends more time and has high potential energy. If you pick a random small interval, one energy will likely dominate the other.

Q: How do we know the average of $\sin^2(\omega t)$ is $1/2$?

You can use the trigonometric identity $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$. When you integrate $\cos(2\omega t)$ over a full period, it evaluates to zero, leaving just the integral of $1/2$, which gives $1/2$.

Q: Is the total energy constant even if there is friction?

No! The total mechanical energy is constant only for *free* or *undamped* Simple Harmonic Motion. If friction or damping is present, the total energy will exponentially decrease over time.

Q: At what exact points are kinetic and potential energies equal?

They are equal when $\sin^2(\omega t) = \cos^2(\omega t)$, which means $\tan^2(\omega t) = 1$. This happens at displacements $x = \pm \frac{a}{\sqrt{2}}$, where $a$ is the amplitude.

Q: Does the mass of the body affect the ratio of average KE to average PE?

Not at all! Both average KE and average PE are directly proportional to the mass $m$. When you take their ratio, the mass cancels out completely, leaving the ratio as exactly 1:1 over a full period.

Question

For a body executing SHM
A. potential energy is always equal to its kinetic energy.
B. average potential and kinetic energy over any given time interval are always equal.
C. sum of the kinetic and potential energy at any point of time is constant.
D. average kinetic energy in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below.

Options:

(A) (C) and (D)

(B) Only (C)

(C) (B) and (C)

(D) Only (B)

Sigma Percentile · Accepted Answer

### The Dance of Energy in Simple Harmonic Motion

Imagine a block attached to a spring, oscillating back and forth on a frictionless surface. This simple setup is the heart of Simple Harmonic Motion (SHM). But beneath this back-and-forth movement lies a beautiful, continuous transformation of energy.

In this problem, we are going to act like energy detectives. We will analyze four different claims about how kinetic and potential energy behave during SHM, and we will use both our mathematical tools and our physical intuition to uncover the truth.

### Analyzing the Energy Equations

Let's start by writing down the fundamental equations. If a particle is executing SHM, its displacement $x$ from the mean position as a function of time $t$ can be written as:

$$x = a \sin(\omega t)$$

where $a$ is the amplitude and $\omega$ is the angular frequency.

The velocity $v$ is the rate of change of displacement:

$$v = \frac{dx}{dt} = a \omega \cos(\omega t)$$

Now, let's construct our energy equations. The potential energy ($PE$) stored in the system is given by $\frac{1}{2} k x^2$. Since $k = m \omega^2$, we can write:

$$PE = \frac{1}{2} m \omega^2 a^2 \sin^2(\omega t)$$

Similarly, the kinetic energy ($KE$) is $\frac{1}{2} m v^2$, which becomes:

$$KE = \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t)$$

Notice the beautiful symmetry here! The potential energy dances to the tune of $\sin^2(\omega t)$, while the kinetic energy follows $\cos^2(\omega t)$.

### Evaluating the Statements

**Statement A: Is PE always equal to KE?**
Look at the equations. One is a sine function, the other is a cosine function. They are out of phase. When the particle is at the mean position ($x = 0$), the potential energy is zero, but the kinetic energy is at its absolute maximum. They only equal each other at specific points (when $x = \pm \frac{a}{\sqrt{2}}$). Therefore, they are **not** always equal. Statement A is false.

**Statement C: The Sum of Energies**
What happens if we add them together at any random instant of time? Let's find the Total Mechanical Energy ($TE$):

$$TE = PE + KE = \frac{1}{2} m \omega^2 a^2 (\sin^2 \omega t + \cos^2 \omega t)$$

Thanks to the most famous trigonometric identity, $\sin^2 	heta + \cos^2 	heta = 1$, the time dependence completely vanishes!

$$TE = \frac{1}{2} m \omega^2 a^2 = 	ext{Constant}$$

The total energy is perfectly conserved. Statement C is absolutely true.

**Statement D: Averages Over a Full Period**
Now, let's talk about averages. To find the average of a function over a full time period $T$, we integrate it from $0$ to $T$ and divide by $T$. 
The average value of both $\sin^2(\omega t)$ and $\cos^2(\omega t)$ over one complete cycle is exactly $\frac{1}{2}$.

Substituting this into our energy equations, we get:

$$PE_{avg} = \frac{1}{2} m \omega^2 a^2 	imes \left(\frac{1}{2}ight) = \frac{1}{4} m \omega^2 a^2$$

$$KE_{avg} = \frac{1}{2} m \omega^2 a^2 	imes \left(\frac{1}{2}ight) = \frac{1}{4} m \omega^2 a^2$$

They are perfectly equal! Over a full time period, the system spends its energy symmetrically. Statement D is true.

**Statement B: Averages Over ANY Interval?**
Here is where many students make a silly mistake. Statement B claims the averages are equal over *any* given time interval.

Imagine taking a tiny time interval right as the particle crosses the mean position. In this brief window, the kinetic energy is huge, and the potential energy is nearly zero. The average KE will massively dominate the average PE in this specific interval. The averages are only guaranteed to be equal over a full period (or a half period), not just *any* random interval. Thus, Statement B is false.

### The Final Verdict

We have rigorously proven that only **Statement C** and **Statement D** are correct. This perfectly matches option (a).

Physics is not just about memorizing formulas; it is about visualizing the interplay of forces and energies. The next time you see a pendulum swing or a spring bounce, remember the invisible, elegant dance of sine and cosine happening right before your eyes!

Select Answer:

Visualized Solution

$E n er g y in S H M$

$K in e t i c an d P o t e n t ia l E n er g y$

$C h ec k in g S t a t e m e n t A$

$C h ec k in g S t a t e m e n t C$

$C h ec k in g S t a t e m e n t D$

$C h ec k in g S t a t e m e n t B$

$F ina l C o n c l u s i o n$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Dance of Energy in Simple Harmonic Motion

Analyzing the Energy Equations

Evaluating the Statements

The Final Verdict

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