Problem Statement
(a) A block of mass $m = 1\,\text{kg}$ is attached to a spring of force constant $k = 130\,\text{N/m}$ on a smooth horizontal surface and set into oscillation with amplitude $A = 0.11\,\text{m}$. Find the angular frequency, the time period, and the maximum speed of the block. (b) A simple pendulum of length $L = 1.25\,\text{m}$ oscillates with small amplitude; find its time period. Take $g = 9.8\,\text{m/s}^{2}$.
Given Information
- $m = 1\,\text{kg}$ — mass of the block
- $k = 130\,\text{N/m}$ — spring constant
- $A = 0.11\,\text{m}$ — amplitude; $L = 1.25\,\text{m}$ — pendulum length
- $g = 9.8\,\text{m/s}^{2}$
Physical Concepts & Formulas
Simple harmonic motion arises whenever the restoring force is proportional to displacement, $F=-kx$. The angular frequency of a spring–mass system is $\sqrt{k/m}$, independent of amplitude. For a simple pendulum with small swings the period depends only on length and $g$. Speed is maximum at the equilibrium position where all energy is kinetic.
- $\omega = \sqrt{\dfrac{k}{m}}$ — spring–mass angular frequency
- $T = \dfrac{2\pi}{\omega}$ — time period
- $v_{\max} = \omega A$ — maximum speed in SHM
- $T_{pend} = 2\pi\sqrt{\dfrac{L}{g}}$ — simple pendulum
Step-by-Step Solution
Step 1 — Angular frequency of the spring–mass system: Use $\omega=\sqrt{k/m}$; it does not depend on the amplitude.
$$ \omega = \sqrt{\frac{130}{1}} = 11.4\,\text{rad/s} $$
Step 2 — Time period of the oscillation: The period is $2\pi$ divided by the angular frequency.
$$ T = \frac{2\pi}{\omega} = 0.551\,\text{s} $$
Step 3 — Maximum speed of the block: Speed peaks at the mean position with magnitude $\omega A$.
$$ v_{\max} = \omega A = 11.4\times0.11 = 1.25\,\text{m/s} $$
Step 4 — Period of the simple pendulum: Apply $T=2\pi\sqrt{L/g}$ for small-amplitude swings.
$$ T_{pend} = 2\pi\sqrt{\frac{1.25}{9.8}} = 2.24\,\text{s} $$
Worked Calculation
$$ \omega = \sqrt{130/1} = 11.4\,\text{rad/s},\ T = 0.551\,\text{s},\ v_{\max} = 1.25\,\text{m/s} $$
Answer
$$ \boxed{\omega = 11.4\,\text{rad/s},\ T = 0.551\,\text{s},\ v_{\max} = 1.25\,\text{m/s}} $$
The period is independent of amplitude (isochronism) and, for the spring, independent of $g$.
Physical Interpretation
Isochronism — the amplitude-independence of the period — is what makes pendulums and springs good timekeepers. A stiffer spring or a lighter mass oscillates faster; a longer pendulum swings more slowly, which is why grandfather clocks are tall.
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