A particle of mass 0.20\text{ kg} executes simple harmonic motion with an amplitude of 0.0… — JAMB UTME

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a period of oscillation of $0.20\pi\text{ s}$ . What is the maximum kinetic energy of the particle in joules?

Answer: 0.025 J

Answer

The maximum kinetic energy of the particle is

0.025\text{ J}

The maximum kinetic energy occurs at the equilibrium position where speed reaches its maximum value

v_{\text{max}} = \omega A

. Substituting

\omega = \frac{2\pi}{0.20\pi} = 10\text{ rad/s}

into the maximum speed formula gives

v_{\text{max}} = 10 \times 0.05 = 0.50\text{ m/s}

. Computing kinetic energy yields

E_k = \frac{1}{2} (0.20) (0.50)^2 = 0.025\text{ J}

Step-by-Step Solution

Calculate angular frequency (\omega)

\omega = 10\text{ rad/s}

Using the relation

\omega = \frac{2\pi}{T}

with

T = 0.20\pi\text{ s}

Calculate maximum speed (v_{max})

v_{max} = 0.50\text{ m/s}

Maximum velocity occurs at the equilibrium position and is given by

v_{\text{max}} = \omega A

Calculate maximum kinetic energy (E_k)

E_k = 0.025\text{ J}

Using the kinetic energy formula

E_k = \frac{1}{2} m v^2

at maximum velocity.

Key Concept

Energy conservation and maximum speed in Simple Harmonic Motion

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