The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 ~joule$0.02 \mathrm{~joule}$. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob = 20 ~g$20 \mathrm{~g}$)

Related Formula

Core Logic

Kinetic energy fluctuates over time as:

Since ²(θ)$\cos^2(\theta)$ produces entirely positive values and cycles twice as fast as (θ)$\cos(\theta)$, the kinetic energy frequency is twice the oscillation frequency of the pendulum. Thus, the period of Kinetic Energy is T/2$T/2$.

Step 1: Graph Identification

The K.E. must be purely positive. The wave should complete two full cycles in the total span of T$T$. It starts from maximum (if pendulum released from mean position, wait, typically graphs assume release from mean if K.E. starts max, or extreme if K.E. starts zero). Option 3 perfectly demonstrates a full cycle returning to identical phase exactly at T/2$T/2$, maintaining entirely positive values.

Pattern Recognition

Kinetic Energy and Potential Energy in SHM always operate at double the frequency of displacement, meaning their time period is exactly half (TKE = T/2$T_{KE} = T/2$).

Chapter Mix

Class 11 Physics: Oscillations

Related Formula

Core Logic

First, calculate the time period T$T$ of the simple pendulum. The time period is the time taken for exactly 1$1$ oscillation.

Step 1: Calculate Length

Using the isolated length formula:

Substitute g = 9.8 ~m/s²$g = 9.8 \mathrm{~m/s}^2$, π² = 9.8$\pi^2 = 9.8$, and T = 2 ~s$T = 2 \mathrm{~s}$:

Pattern Recognition

A simple pendulum with a time period of exactly 2 ~s$2 \mathrm{~s}$ is known as a 'seconds pendulum'. The length of a seconds pendulum on Earth is always approximately 1 ~m$1 \mathrm{~m}$.

Chapter Mix

Class 11 Physics: Oscillations