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Question

A mass $m$ is attached to a spring with spring constant $k$ and oscillates on a smooth horizontal surface.  At what displacement from equilibrium is the kinetic energy equal to the potential energy?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option C
## CONCEPT FRAMEWORK
	**Primary Concept**: Simple Harmonic Motion (SHM) and Energy Conservation
	**Related Topics**: Kinetic Energy, Potential Energy, Amplitude of SHM
	**Prerequisites**: Understanding of energy in SHM, formulas for kinetic and potential energy of a spring.
## SOLUTION PATHWAY
**Given Information**:
	* Mass $m$ attached to a spring with spring constant $k$.
	* Oscillating on a smooth horizontal surface.
	* Kinetic Energy (KE) = Potential Energy (PE)
	* Let A be the amplitude
**Method & Steps**:
	1.  Write the expressions for Kinetic Energy (KE) and Potential Energy (PE) in SHM.
		Reasoning: In SHM, the total energy is constant, and it is constantly converted between KE and PE.
		Formula/Rule: $KE = \frac{1}{2}mv^2$, $PE = \frac{1}{2}kx^2$, Total Energy $E = \frac{1}{2}kA^2$ where A is the amplitude.
	2.  Relate velocity to displacement.
		Reasoning: In SHM, $v = \omega \sqrt{A^2 - x^2}$ where $\omega = \sqrt{k/m}$ 
		Formula/Rule: $KE = \frac{1}{2}m\omega^2(A^2-x^2) = \frac{1}{2}k(A^2 - x^2)$
	3.  Set KE equal to PE and solve for x.
		Working: $\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$, so $A^2 - x^2 = x^2$ giving $2x^2 = A^2$ and $x = \frac{A}{\sqrt{2}}$
		Key Point: The total energy is also equal to the maximum potential energy.
	4. Final step: Find the displacement.
		Result: At a displacement of $\frac{A}{\sqrt{2}}$ from equilibrium, the kinetic energy equals the potential energy.
		Verification: When $x = \frac{A}{\sqrt{2}}$, KE = PE = $\frac{1}{4}kA^2$. Total Energy = $\frac{1}{2}kA^2$. 2*(PE or KE) = Total Energy
 ## OPTION ANALYSIS
[A] :x: $A$ is the amplitude where all energy is potential, and KE is zero.
[B] :x: At $A/2$, the potential energy is $1/4$ of the total energy, and kinetic energy is $3/4$ of the total energy.
[C] ✓: At $A/\sqrt{2}$, the kinetic energy equals the potential energy.
[D] :x: At $A/4$, the potential energy is 1/16 of the total, and KE is 15/16 of the total energy.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.  [Using incorrect formulas for KE and PE]: [Remember the correct formulas for spring potential energy and kinetic energy in SHM]
		2. [Not relating velocity to displacement in SHM]: [Remember the velocity displacement relationship for SHM]
	**Quick Tips**:
		1. [Remember that at equilibrium, PE=0 and KE is maximum, and at maximum displacement, KE=0 and PE is maximum] 
		2.  [A useful trick is to equate KE to half the total energy and solve for x. PE is also half the total energy at that point.]

Mr. ANIL KUMAR.G · Answer

$A$

Mr. ANIL KUMAR.G · Answer

$A/2$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$A/\sqrt{2}$"

Mr. ANIL KUMAR.G · Answer

$A/4$

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