When the displacement of a spring is reduced to half its original value, and the spring constant remains uncha

Question

When the displacement of a spring is reduced to half its original value, and the spring constant remains unchanged, what happens to the spring's potential energy?

Examoogle · Accepted Answer

The correct answer is Option B: It decreases to 1/4 of its original value. The potential energy stored in a spring is directly proportional to the square of its displacement. When the displacement is halved, the energy becomes one-fourth of the original value due to the square relationship.

Step-by-Step Solution & Concepts:
The potential energy stored in a spring is directly proportional to the square of its displacement. When the displacement is halved, the energy becomes one-fourth of the original value due to the square relationship.

Working Principle: The elastic potential energy ($U$) stored in a spring is calculated using the formula $U = \frac{1}{2} k x^2$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. Since $U \propto x^2$, if the new displacement $x' = \frac{x}{2}$, then the new potential energy $U' = \frac{1}{2} k (\frac{x}{2})^2 = \frac{1}{2} k (\frac{x^2}{4}) = \frac{1}{4} U$.

Key Formulas: $U = \frac{1}{2} k x^2$ — Elastic potential energy of a spring; $U \propto x^2$ — Relationship between energy and displacement

Key Points: Potential energy is a scalar quantity.; The spring constant $k$ depends on the material and geometry of the spring.; The displacement $x$ is measured from the natural, unstretched position.

Advantages: Useful for calculating work done in spring-mass systems.; Critical for understanding simple harmonic motion (SHM).

Applications: Mechanical watch movements; Vehicle suspension systems

Additional Info: The SI unit for potential energy is Joule ($J$).. Option A is incorrect because it ignores the quadratic dependency on displacement.. Option C is incorrect as it assumes a linear inverse relationship rather than a square.. Option D is incorrect as potential energy changes whenever the state of the spring changes.

Answer: B is correct — Because the potential energy is proportional to the square of the displacement, halving the displacement results in a reduction to $1/4$ of the original energy.

Exam Tip: In any physics problem involving a 'square' relationship (like $E = \frac{1}{2}mv^2$ or $U = \frac{1}{2}kx^2$), always check the power of the variable before calculating the change.

Examoogle · Answer

Option A: It decreases to 1/2 of its original value

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Option C: It increases to 2 times its original value

Examoogle · Answer

Option D: It remains unchanged

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