A body of mass 0.50\text{ kg} connected to a light helical spring of force constant 32\tex… — JAMB UTME

A body of mass $0.50\text{ kg}$ connected to a light helical spring of force constant $32\text{ N/m}$ executes simple harmonic motion on a smooth horizontal surface. If the total mechanical energy of the oscillating system is $0.16\text{ J}$ , what is the maximum speed of the body in $\text{m/s}$ ?

Answer: 0.8 m/s

Answer

The maximum speed of the body is

0.8\text{ m/s}

The total mechanical energy in simple harmonic motion is equal to the maximum kinetic energy at the equilibrium position:

E = \frac{1}{2} m v_{\text{max}}^2

. Substituting

E = 0.16\text{ J}

and

m = 0.50\text{ kg}

yields

0.16 = 0.25 v_{\text{max}}^2

, so

v_{\text{max}}^2 = 0.64

and

v_{\text{max}} = 0.8\text{ m/s}

Step-by-Step Solution

Relate total energy to maximum kinetic energy

E = \frac{1}{2} m v_{\text{max}}^2

At the equilibrium position, potential energy is zero and total mechanical energy is entirely kinetic.

Substitute given values into the equation

0.16 = \frac{1}{2} (0.50) v_{\text{max}}^2

Mass

m = 0.50\text{ kg}

and total energy

E = 0.16\text{ J}

are provided.

Solve for the maximum speed

v_{\text{max}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.8\text{ m/s}

Solving for

v_{\text{max}}

gives

0.8\text{ m/s}

Key Concept

Conservation of Energy in Simple Harmonic Motion

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