HC Verma Chapter 12 Problem 4

Problem Statement

(a) A block of mass $m = 1.5\,\text{kg}$ is attached to a spring of force constant $k = 90\,\text{N/m}$ on a smooth horizontal surface and set into oscillation with amplitude $A = 0.13\,\text{m}$. Find the angular frequency, the time period, and the maximum speed of the block. (b) A simple pendulum of length $L = 1.5\,\text{m}$ oscillates with small amplitude; find its time period. Take $g = 9.8\,\text{m/s}^{2}$.

Given Information

  • $m = 1.5\,\text{kg}$ — mass of the block
  • $k = 90\,\text{N/m}$ — spring constant
  • $A = 0.13\,\text{m}$ — amplitude; $L = 1.5\,\text{m}$ — 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.

$$ \omega = \sqrt{\frac{90}{1.5}} = 7.75\,\text{rad/s} $$

Step 2 — Time period of the oscillation: The period is $2\pi$ divided by the angular frequency.

$$ T = \frac{2\pi}{\omega} = 0.811\,\text{s} $$

Step 3 — Maximum speed of the block: Speed peaks at the mean position with magnitude $\omega A$.

$$ v_{\max} = \omega A = 7.75\times0.13 = 1.01\,\text{m/s} $$

Step 4 — Period of the simple pendulum: Apply $T=2\pi\sqrt{L/g}$ for small-amplitude swings.

$$ T_{pend} = 2\pi\sqrt{\frac{1.5}{9.8}} = 2.46\,\text{s} $$

Worked Calculation

$$ \omega = \sqrt{90/1.5} = 7.75\,\text{rad/s},\ T = 0.811\,\text{s},\ v_{\max} = 1.01\,\text{m/s} $$

Answer

$$ \boxed{\omega = 7.75\,\text{rad/s},\ T = 0.811\,\text{s},\ v_{\max} = 1.01\,\text{m/s}} $$

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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