A body of mass 0.4\text{ kg} suspended vertically from a helical spring produces a static … — JAMB UTME

A body of mass $0.4\text{ kg}$ suspended vertically from a helical spring produces a static extension of $0.1\text{ m}$ . The body is then pulled down further and set into vertical simple harmonic motion with an amplitude of $0.05\text{ m}$ . What is the maximum velocity of the body in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Answer: 0.5 m/s

Answer

The maximum velocity of the body during oscillation is

0.5\text{ m/s}

At static equilibrium, weight balances restoring force (

mg = ke

), giving

\frac{k}{m} = \frac{g}{e} = \frac{10}{0.1} = 100\text{ s}^{-2}

. The angular frequency is

\omega = \sqrt{\frac{k}{m}} = 10\text{ rad/s}

. In SHM, the maximum velocity occurs at the central equilibrium position and is given by

v_{\max} = \omega A = 10 \times 0.05 = 0.5\text{ m/s}

Step-by-Step Solution

Relate spring stiffness to static extension

\frac{k}{m} = 100\text{ s}^{-2}

At vertical static equilibrium, the weight of the mass equals the restoring force:

mg = ke \implies \frac{k}{m} = \frac{g}{e} = \frac{10\text{ m/s}^2}{0.1\text{ m}} = 100\text{ s}^{-2}

Determine the angular frequency

\omega = 10\text{ rad/s}

The angular frequency of a mass-spring system is given by

\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10\text{ rad/s}

Calculate maximum velocity

v_{\max} = 0.5\text{ m/s}

The maximum speed in simple harmonic motion occurs at the equilibrium position and is computed using

v_{\max} = \omega A = 10\text{ rad/s} \times 0.05\text{ m} = 0.5\text{ m/s}

Key Concept

Maximum velocity and angular frequency derived from static extension in Simple Harmonic Motion

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