Question

The bob of a simple pendulum of length 'L' is released from a position of small angular displacement $\theta$. Its linear displacement at time 't' is ($g =$ acceleration due to gravity)

• A. $L\theta \cos \left[ \sqrt{\frac{g}{L}} \cdot t \right]$
• B. $L\theta \sin \left[ 2\pi\sqrt{\frac{g}{L}} \cdot t \right]$
• C. $L\theta \cos \left[ 2\pi\sqrt{\frac{g}{L}} \cdot t \right]$
• D. $L\theta \sin \left[ \sqrt{\frac{g}{L}} \cdot t \right]$

Hint:

Physics Tip : The $\cos$ function is used here because the motion begins at the maximum displacement (the extreme position) at $t=0$.

Answer 1

The Correct Option is A

Concept: Physics (Oscillations) - Displacement Equation for Simple Harmonic Motion (S.H.M.).

Step 1: State the standard displacement equation. For a particle performing S.H.M. released from an extreme position, the angular displacement $y$ at time $t$ is given by: $$y = A \cos(\omega t)$$ where $A$ is the amplitude and $\omega$ is the angular frequency.

Step 2: Determine the angular frequency of the pendulum. The time period $T$ of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$. Since $\omega = \frac{2\pi}{T}$, we have: $$\omega = \sqrt{\frac{g}{L}}$$

Step 3: Substitute amplitude and frequency into the equation. For small angular displacement $\theta$, the linear amplitude is $A = L\theta$. Substituting $A$ and $\omega$: $$s = L\theta \cos \left[ \sqrt{\frac{g}{L}} \cdot t \right]$$ $$ \therefore \text{The linear displacement is } L\theta \cos \left[ \sqrt{\frac{g}{L}} \cdot t \right]. \text{} $$