A particle starts executing simple harmonic motion (SHM) of amplitude a and total energy E. At any i

Q: Why is the total energy $E$ equal to $\frac{1}{2} K a^2$?

In SHM, the total mechanical energy is conserved. At the extreme position ($y = a$), the velocity is zero, so kinetic energy is zero. All the energy is stored as potential energy, which is $\frac{1}{2} K y^2$. Substituting $y = a$ gives the total energy $E = \frac{1}{2} K a^2$.

Q: Can we solve this using the velocity equation $v = \omega \sqrt{a^2 - y^2}$?

Absolutely! Kinetic energy is $\frac{1}{2} m v^2$. Substituting $v^2 = \omega^2 (a^2 - y^2)$ gives $KE = \frac{1}{2} m \omega^2 (a^2 - y^2)$. Since $K = m \omega^2$, this is exactly the same as $KE = \frac{1}{2} K (a^2 - y^2)$.

Q: What happens to the potential energy at this displacement?

Since total energy is conserved ($E = KE + PE$), and $KE = \frac{3E}{4}$, the potential energy must be $PE = E - \frac{3E}{4} = \frac{E}{4}$. This means at $y = \frac{a}{2}$, $25\%$ of the energy is potential and $75\%$ is kinetic.

Q: Why did we ignore the negative root when taking the square root of $y^2 = \frac{a^2}{4}$?

Mathematically, $y = \pm \frac{a}{2}$. The negative sign simply means the particle could be at a displacement $\frac{a}{2}$ on the opposite side of the mean position. Since the options only provide the positive magnitude, we select $y = \frac{a}{2}$.

Q: Is the kinetic energy half of the total energy at $y = \frac{a}{2}$?

No, that's a common misconception! Because energy depends on the *square* of the displacement, at half the amplitude ($y = \frac{a}{2}$), the potential energy is only $\frac{1}{4}$ of the total, leaving $\frac{3}{4}$ for kinetic energy. The energies are equal ($50\%$ each) at $y = \frac{a}{\sqrt{2}}$.

Question

A particle starts executing simple harmonic motion (SHM) of amplitude $a$ and total energy $E$. At any instant, its kinetic energy is $\frac{3E}{4}$, then its displacement $y$ is given by

Options:

(A) $y = a$

(B) $y = \frac{a}{\sqrt{2}}$

(C) $y = \frac{a\sqrt{3}}{2}$

(D) $y = \frac{a}{2}$

Sigma Percentile · Accepted Answer

The problem asks us to find the displacement of a particle executing Simple Harmonic Motion (SHM) when its kinetic energy is exactly $75\%$ (or $\frac{3}{4}$) of its total energy. Let's break this down step-by-step and understand the beautiful interplay between kinetic and potential energy in SHM.

### Analyzing the Setup
Imagine a particle, like a block attached to a spring, oscillating back and forth on a frictionless surface. It moves from its mean position ($y = 0$) to a maximum displacement, which we call the amplitude, denoted by $a$.

At this extreme position ($y = a$), the particle momentarily stops before reversing its direction. Because its velocity is zero, its kinetic energy is zero, and all its energy is stored as potential energy. This maximum potential energy is equal to the total mechanical energy of the system.

### The Master Equations
The total mechanical energy $E$ of a particle in simple harmonic motion remains constant throughout its journey. It depends only on the spring constant $K$ and the square of the amplitude $a$. We can express this mathematically as:
$$E = \frac{1}{2} K a^2$$

Now, as the particle moves from the extreme position towards the mean position, it speeds up. Its potential energy converts into kinetic energy. At any random displacement $y$, the kinetic energy $KE$ is given by the formula:
$$KE = \frac{1}{2} K (a^2 - y^2)$$

Notice how this equation perfectly describes the physics: when $y = 0$ (mean position), $KE$ is maximum ($\frac{1}{2} K a^2$), and when $y = a$ (extreme position), $KE$ is zero.

### Applying the Given Condition
The question gives us a very specific scenario. It states that at a particular displacement $y$, the kinetic energy is exactly three-fourths of the total energy $E$. 
$$KE = \frac{3E}{4}$$

Let's set up our equation by substituting this given value into our kinetic energy formula:
$$\frac{3E}{4} = \frac{1}{2} K (a^2 - y^2)$$

Here is where the magic happens. Remember our master equation for total energy $E$? Let's plug that right into our new equation. We replace $E$ with $\frac{1}{2} K a^2$. Now we have an equation entirely in terms of $K$, $a$, and $y$:
$$\frac{3}{4} \left( \frac{1}{2} K a^2 \right) = \frac{1}{2} K (a^2 - y^2)$$

### Final Calculation
Look closely at both sides of the equation. We have the term $\frac{1}{2} K$ common to both. Let's cancel it out. This simplifies our equation beautifully:
$$\frac{3}{4} a^2 = a^2 - y^2$$

The physics is done; now it's just simple algebra! Let's isolate our unknown variable, $y$. We move $y^2$ to the left side and $\frac{3}{4} a^2$ to the right:
$$y^2 = a^2 - \frac{3}{4} a^2$$

Subtracting three-fourths from one whole leaves us with exactly one-fourth:
$$y^2 = \frac{a^2}{4}$$

We are at the final step. To find the displacement $y$, we simply take the square root of both sides:
$$y = \frac{a}{2}$$

So, the particle is exactly halfway between the mean position and the extreme position when its kinetic energy is $75\%$ of the total energy.

**Physical Intuition Check:** If the kinetic energy is $75\%$ of the total energy, the remaining $25\%$ must be potential energy. Since potential energy is proportional to $y^2$, and $y = \frac{a}{2}$, $y^2$ is indeed $\frac{1}{4}$ of $a^2$. Everything aligns perfectly!

Animated Solution for Physics - Oscillations: A particle starts executing simple harmonic motion (SHM) of amplitude $a$ and total energy $E$ . At any instant, its kinetic energy is $\frac{3 E}{4}$ , then its displacement $y$ is given by

Select Answer:

Visualized Solution

System Setup

Total Energy of SHM

Kinetic Energy at Displacement $y$

Applying the Given Condition

Substituting Total Energy

Simplifying the Equation

Solving for $y^{2}$

Final Displacement

The Way Forward

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup Imagine a particle, like a block attached to a spring, oscillating back and forth on a frictionless surface

The Master Equations The total mechanical energy $E$ of a particle in simple harmonic motion remains constant throughout its journey

Applying the Given Condition The question gives us a very specific scenario

Final Calculation Look closely at both sides of the equation

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