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Chapter 11: Problem 24

The potential energy of a particle executing S.H.M. at a distance \(x\) from the mean position is proportional to [Roorkee 1992] (a) \(\sqrt{x}\) (b) \(x\) (c) \(x^{2}\) (d) \(x^{3}\)

Short Answer

Expert verified
The potential energy of a particle executing S.H.M. at a distance \(x\) from the mean position is proportional to \(x^{2}\).

Step by step solution

01

Stating the general formula for P.E. in S.H.M.

The potential energy (P.E.) in Simple Harmonic Motion (S.H.M.) is given by the equation: \(P.E. = \frac{1}{2} kx^{2}\), where \(k\), known as the spring constant, measures the restoring force per unit displacement and \(x\) is the displacement from the mean or equilibrium position.
02

Analyzing the form of the equation

Upon inspecting this equation, it's evident that the potential energy is proportional to the square of ''x'', the displacement from the equilibrium position.
03

Comparing with the given options

Therefore, the potential energy is not proportional to \(\sqrt{x}\), \(x\), or \(x^{3}\). Hence the correct answer is that P.E. is proportional to \(x^{2}\).

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Most popular questions from this chapter

A simple harmonic oscillator has an amplitude \(A\) and time period \(T\). The time required by it to travel from \(x=A\) to \(x=A / 2\) is \(\quad\) [CBSE 1992; SCRA 1996] (a) \(T / 6\) (b) \(T / 4\) (c) \(T / 3\) (d) \(T / 2\)

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