A block whose mass m = 4 kg is fastened to a spring with a spring constant k = 64 N/m. The block is pulled from its equilibrium position on a frictionless surface and released. The period of the resulting motion in seconds is

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Question

This question was previously asked in

ISRO VSSC Technical Assistant Mechanical 9 June 2019 Official Paper

Answer 1

Concept:

In Simple harmonic motion, we have

Time period, \(T = 2π \sqrt {\frac{m}{k}}\)

where, T = Time period of oscillation, m = mass of the block, k = Spring constant

Calculation:

Given:

We have, m = 4 kg, k = 64 N/m

Since, \(T = 2π \sqrt {\frac{m}{k}}\) placing the given values we get,

\(T = 2π \sqrt {\frac{4}{{64}}} \)

T = π/2

Hence, the period of the resulting motion in seconds is T = π/2.