A body of mass 0.40\text{ kg} is attached to a light helical spring and set into simple ha… — JAMB UTME

A body of mass $0.40\text{ kg}$ is attached to a light helical spring and set into simple harmonic motion. If the system oscillates with a period of $0.40\pi\text{ s}$ , what is the force constant of the spring in $\text{N/m}$ ?

Answer: 10 N/m

Answer

The force constant of the spring is

10\text{ N/m}

The period of a mass-spring system undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{m}{k}}

. Substituting

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

yields

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

. Dividing both sides by

2\pi

gives

0.20 = \sqrt{\frac{0.40}{k}}

. Squaring both sides yields

0.04 = \frac{0.40}{k}

, which gives

k = 10\text{ N/m}

Step-by-Step Solution

Identify the period relationship for a spring-mass SHM system

T = 2\pi \sqrt{\frac{m}{k}}

This formula relates the period of oscillation

T

to the mass

m

and spring constant

k

Substitute the given mass and period into the equation

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

Given that

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

Simplify the equation by isolating the square root term

\sqrt{\frac{0.40}{k}} = 0.20

Dividing both sides of the equation by

2\pi

isolates the radical.

Square both sides and solve for the spring constant

k

k = 10\text{ N/m}

Squaring yields

0.04 = \frac{0.40}{k}

, which rearranges to

k = \frac{0.40}{0.04} = 10\text{ N/m}

Key Concept

Period of oscillation of a mass-spring system in Simple Harmonic Motion

Estimated Time:1m 30s

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