For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:

Q: For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by:

Kinetic energy fluctuates over time as: Since 2(theta) produces entirely positive values and cycles twice as fast as (theta), the kinetic energy frequency is twice the oscillation frequency of the pendulum. Thus, the period of Kinetic Energy is T/2. The K.E. must be purely positive. The wave should complete two full cycles in the total span of 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.

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2024

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Questions

Solution & Explanation

Related Formula

x(t) = A (ω t + φ)

v(t) = Aω (ω t + φ)

K.E. = (1)/(2) m v²

Core Logic

Kinetic energy fluctuates over time as:

K.E. = (1)/(2) m A² ω² ²(ω t + φ)

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

More Oscillations Previous-Year Questions

Q32 neet_2026_03_may_morning Simple Pendulum

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum

L

, notes down the data of time taken to complete

30

oscillations as

60 ~s

and hence calculates the length of the simple pendulum as : (Take

π² = 9.8

, and

g = 9.8 ~m/s²

)

A. $2 ~m$ $2 \mathrm{~m}$
B. $0.75 ~m$ $0.75 \mathrm{~m}$
C. $1.5 ~m$ $1.5 \mathrm{~m}$
D. $1 ~m$ $1 \mathrm{~m}$

Solution

Related Formula

T = 2π \sqrt(( )/(g))

= (g T²)/(4π²)

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.

T = Total TimeNumber of Oscillations = 60 ~s30 = 2 ~s

Step 1: Calculate Length

Using the isolated length formula:

= (g T²)/(4π²)

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

= (9.8 \times (2)²)/(4 \times 9.8)

= (9.8 \times 4)/(4 \times 9.8) = 1 ~m

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

Q36 neet_2026_03_may_morning Energy in Simple Harmonic Motion

The sum of kinetic energy and potential energy of a simple pendulum bob is

0.02 ~joule

. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob =

20 ~g

)

A. $1.41 ~m/s$ $1.41 \mathrm{~m/s}$
B. $14.1 ~m/s$ $14.1 \mathrm{~m/s}$
C. $0.2 ~m/s$ $0.2 \mathrm{~m/s}$
D. $2.0 ~m/s$ $2.0 \mathrm{~m/s}$

Solution

Related Formula

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

KE = (1)/(2) m v²

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.

KEmax = 0.02 ~J

Step 1: Calculate Speed

(1)/(2) m v² = 0.02

Given mass $m = 20 ~g = 20 \times 10⁻³ ~kg = 0.02 ~kg$ .

(1)/(2) (0.02) v² = 0.02

(1)/(2) v² = 1

$v² = 2$ $v^2 = 2$

v = \sqrt(2) \approx 1.41 ~m/s

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 $\sqrt(2) \approx 1.414$ $\sqrt{2} \approx 1.414$ immediately.

Chapter Mix

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

For a simple pendulum, having time period $T$ $T$, the variation of kinetic energy (K.E.) with time ( $t$ $t$) is represented by:

Solution & Explanation

Related Formula

Core Logic

Step 1: Graph Identification

Pattern Recognition

Chapter Mix

Reference Study Guides

More Oscillations Previous-Year Questions

Solution

Related Formula

Core Logic

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Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Speed

Pattern Recognition

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For a simple pendulum, having time period T$T$, the variation of kinetic energy (K.E.) with time (t$t$) is represented by:

Solution & Explanation

Related Formula

Core Logic

Step 1: Graph Identification

Pattern Recognition

Chapter Mix

Reference Study Guides

More Oscillations Previous-Year Questions

Solution

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Core Logic

Step 1: Calculate Length

Pattern Recognition

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Pattern Recognition

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For a simple pendulum, having time period $T$ $T$, the variation of kinetic energy (K.E.) with time ( $t$ $t$) is represented by: