The Question is from NEET guide , and it was created by the author VENKATESWARARAO

Question

A spring with spring constant $k$ is stretched by a distance $x$. What is the maximum speed of a mass $m$ attached to the spring during its oscillation?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Simple Harmonic Motion (SHM) and Conservation of Energy
	**Related Topics**: Spring potential energy, kinetic energy
	**Prerequisites**: Understanding of SHM, energy conservation
## SOLUTION PATHWAY
**Given Information**:
		Spring constant = $k$
		Displacement = $x$
		Mass = $m$
		Important conditions: The system is undergoing simple harmonic motion.
**Method & Steps**:
	1. Apply the principle of conservation of energy.
		Reasoning: At maximum displacement ($x$), the potential energy is maximum and kinetic energy is zero. At maximum speed, the kinetic energy is maximum and potential energy is zero.
		Formula/Rule: $PE_{spring} = \frac{1}{2}kx^2$, $KE = \frac{1}{2}mv^2$
	2. Equate the maximum potential energy to the maximum kinetic energy.
		Working: $\frac{1}{2}kx^2 = \frac{1}{2}mv_{max}^2$
		Key Point: The maximum speed occurs when the mass passes through the equilibrium position (x=0).
	3. Solve for the maximum speed ($v_{max}$).
		Result: $v_{max} = x\sqrt{\frac{k}{m}}$
		Verification: The units are consistent: $[v_{max}] = \sqrt{\frac{[k]}{[m]}}[x] = \sqrt{\frac{kg \cdot m \cdot s^{-2}}{kg}} \cdot m = m \cdot s^{-1}$ which is the unit of velocity.
## OPTION ANALYSIS
[A] :x: $\frac{kx}{m}$: This has units of acceleration, not velocity.
[B] ✓: $x\sqrt{\frac{k}{m}}$: This is the correct expression for maximum velocity derived from energy conservation.
[C] :x: $x\sqrt{km}$: This expression has incorrect units and does not account for energy conservation.
[D] :x: $\frac{kx}{\sqrt{m}}$: This expression has incorrect units and does not account for energy conservation.
## LEARNING AIDS
	**Common Mistakes**:
		1.Confusing potential energy with kinetic energy: Remember that at maximum displacement, potential energy is maximum and kinetic energy is zero, and vice-versa.
		2.Incorrectly applying the formula for energy: Ensure correct substitution of values and units in the energy conservation equation.
	**Quick Tips**:
		1.Use the principle of conservation of energy to solve SHM problems.
		2.Remember that the maximum speed in SHM occurs at the equilibrium position.

VENKATESWARARAO · Answer

$\frac{kx}{m}$

VENKATESWARARAO · Answer

The correct Option is "$x\sqrt{\frac{k}{m}}$"

VENKATESWARARAO · Answer

$x\sqrt{km}$

VENKATESWARARAO · Answer

$\frac{kx}{\sqrt{m}}$

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