A particle executes simple harmonic motion along a straight line with an amplitude of 0.10… — JAMB UTME

A particle executes simple harmonic motion along a straight line with an amplitude of $0.10\text{ m}$ . At what displacement from the equilibrium position is the kinetic energy of the particle equal to three times its potential energy?

$0.05\text{ m}$ Answer
B
$0.087\text{ m}$
C
$0.071\text{ m}$
D
$0.025\text{ m}$

Answer

The displacement from the equilibrium position is

0.05\text{ m}

In simple harmonic motion, potential energy is

U = \frac{1}{2}kx^2

and kinetic energy is

K = \frac{1}{2}k(A^2 - x^2)

. Equating

K = 3U

yields

A^2 - x^2 = 3x^2

, which simplifies to

4x^2 = A^2

x = \frac{A}{2}

. For an amplitude of

0.10\text{ m}

, the displacement is

0.05\text{ m}

Step-by-Step Solution

Write the expressions for kinetic energy

K

and potential energy

U

in simple harmonic motion.

U = \frac{1}{2} k x^2

and

K = \frac{1}{2} k (A^2 - x^2)

, where

A

is amplitude and

x

is displacement.

These equations express the energy distribution at any displacement

x

Set up the condition given in the problem,

K = 3U

\frac{1}{2} k (A^2 - x^2) = 3 \times \left(\frac{1}{2} k x^2\right) \implies A^2 - x^2 = 3x^2

Canceling common factor

\frac{1}{2} k

simplifies the relationship between amplitude and displacement.

Solve for displacement

x

in terms of amplitude

A

A^2 = 4x^2 \implies x = \frac{A}{2}

Taking the square root of both sides gives the position where kinetic energy is three times potential energy.

Substitute the given amplitude

A = 0.10\text{ m}

to find

x

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

Carrying out the calculation yields the final numerical displacement.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Estimated Time:1m 15s

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