A vertical light spring stretches by 0.10\text{ m} when a block is suspended from it in eq… — JAMB UTME

A vertical light spring stretches by $0.10\text{ m}$ when a block is suspended from it in equilibrium. The block is then pulled down an additional $0.05\text{ m}$ from its equilibrium position and released from rest to undergo simple harmonic motion. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the block when it is at a displacement of $0.03\text{ m}$ from its equilibrium position?

$0.40\text{ m/s}$ Answer
B
$0.50\text{ m/s}$
C
$0.30\text{ m/s}$
D
$0.04\text{ m/s}$

Answer

0.40\text{ m/s}

The system's angular frequency

\omega

is determined by the equilibrium extension

e

using

\omega = \sqrt{\frac{g}{e}} = 10\text{ rad/s}

. Combining this with the amplitude

A = 0.05\text{ m}

in the SHM speed relation

v = \omega \sqrt{A^2 - y^2}

y = 0.03\text{ m}

yields

v = 10 \sqrt{(0.05)^2 - (0.03)^2} = 0.40\text{ m/s}

Step-by-Step Solution

Calculate the angular frequency of the mass-spring system using static equilibrium conditions.

\omega = 10\text{ rad/s}

At equilibrium,

mg = ke \implies \frac{k}{m} = \frac{g}{e}

. Therefore,

\omega = \sqrt{\frac{g}{e}} = \sqrt{\frac{10}{0.10}} = 10\text{ rad/s}

Identify the amplitude of simple harmonic motion.

A = 0.05\text{ m}

The initial displacement from the equilibrium position when released from rest defines the amplitude of oscillation.

Apply the SHM velocity-displacement formula

v = \omega \sqrt{A^2 - y^2}

y = 0.03\text{ m}

v = 0.40\text{ m/s}

v = 10 \times \sqrt{(0.05)^2 - (0.03)^2} = 10 \times \sqrt{0.0025 - 0.0009} = 10 \times 0.04 = 0.40\text{ m/s}

Key Concept

Relating static extension to angular frequency and calculating instantaneous speed in Simple Harmonic Motion

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