Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L$L$, notes down the data of time taken to complete 30$30$ oscillations as 60 ~s$60 \mathrm{~s}$ and hence calculates the length of the simple pendulum as : (Take π² = 9.8$\pi^2 = 9.8$, and g = 9.8 ~m/s²$\mathrm{g} = 9.8 \mathrm{~m/s}^2$)

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

TE = KE + PE$TE = KE + PE$ At equilibrium, PE = 0 TE = KEmax$PE = 0 \implies TE = KE_{\text{max}}$

Core Logic

The sum of kinetic and potential energy is the Total Energy (TE), which remains constant. Thus, TE = 0.02 ~J$TE = 0.02 \mathrm{~J}$. At the equilibrium (mean) position, the potential energy is zero, meaning all the total energy is converted entirely into kinetic energy.

Step 1: Calculate Speed

Given mass m = 20 ~g = 20 × 10⁻³ ~kg = 0.02 ~kg$m = 20 \mathrm{~g} = 20 \times 10^{-3} \mathrm{~kg} = 0.02 \mathrm{~kg}$.

v² = 2$v^2 = 2$

Pattern Recognition

Mass and energy are intentionally matched (0.02 J and 0.02 kg) to instantly reduce the algebraic equation to v² = 2$v^2 = 2$. Recognise √(2) ≈ 1.414$\sqrt{2} \approx 1.414$ immediately.

Chapter Mix

Class 11 Physics: Oscillations Class 11 Physics: Work Energy and Power