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Oscillations - Simple Pendulum

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An ideal simple pendulum consists of a point mass (bob) suspended by a massless, inextensible string of length LL from a frictionless rigid support.

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The restoring force acting on the bob is F=−mgsin⁡θF = -mg \sin \theta. For small angular displacements, sin⁡θ≈θ\sin \theta \approx \theta (in radians), making the motion Simple Harmonic Motion (SHM).

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The differential equation of motion for a simple pendulum is given by d2θdt2+gLθ=0\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0, where gg is the acceleration due to gravity.

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The time period TT is independent of the mass of the bob and the amplitude of oscillation (provided the amplitude is small).

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A 'Seconds Pendulum' is defined as a pendulum that has a time period of exactly 2 s2\text{ s}, meaning it takes 1 s1\text{ s} to travel from one extreme position to the other.

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If the pendulum is placed in an accelerating frame (like a lift), the effective acceleration due to gravity geffg_{eff} changes, thereby altering the time period T=2πLgeffT = 2\pi \sqrt{\frac{L}{g_{eff}}}.

📐Formulae

F=−mgsin⁡θF = -mg \sin \theta

ω=gL\omega = \sqrt{\frac{g}{L}}

T=2πLgT = 2\pi \sqrt{\frac{L}{g}}

f=12πgLf = \frac{1}{2\pi} \sqrt{\frac{g}{L}}

Lseconds=gT24π2≈9.8×224×(3.14)2≈0.994 mL_{seconds} = \frac{g T^2}{4\pi^2} \approx \frac{9.8 \times 2^2}{4 \times (3.14)^2} \approx 0.994\text{ m}

💡Examples

Problem 1:

Calculate the time period of a simple pendulum of length 1.0 m1.0\text{ m} on the surface of the Moon, where gmoon=1.6 m/s2g_{moon} = 1.6\text{ m/s}^2.

Solution:

T=2πLgmoonT = 2\pi \sqrt{\frac{L}{g_{moon}}} T=2×3.14×1.01.6T = 2 \times 3.14 \times \sqrt{\frac{1.0}{1.6}} T≈6.28×0.625≈6.28×0.79≈4.96 sT \approx 6.28 \times \sqrt{0.625} \approx 6.28 \times 0.79 \approx 4.96\text{ s}

Explanation:

The time period is calculated using the standard formula by substituting the local acceleration due to gravity on the Moon. Since gg is lower on the Moon, the time period is significantly longer than on Earth.

Problem 2:

If the length of a simple pendulum is increased by 21%21\%, what is the percentage increase in its time period?

Solution:

Let the initial length be L1L_1 and final length be L2=1.21L1L_2 = 1.21 L_1. Since T∝LT \propto \sqrt{L}: T2T1=L2L1=1.21=1.1\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}} = \sqrt{1.21} = 1.1 Percentage increase = (1.1−1)×100%=10%(1.1 - 1) \times 100\% = 10\%

Explanation:

Because the time period is proportional to the square root of the length, a 21%21\% increase in length leads to a 10%10\% increase in the time period (1.12=1.211.1^2 = 1.21).

Simple Pendulum - Revision Notes & Key Formulas | CBSE Class 11 Physics