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Oscillations - Simple Harmonic Motion (SHM)

Grade 11CBSEPhysics

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

🔑Concepts

Periodic Motion: A motion that repeats itself at regular intervals of time. Oscillatory motion is a specific type of periodic motion where the body moves to and fro about a fixed mean position.

Simple Harmonic Motion (SHM): A type of oscillation where the restoring force is directly proportional to the displacement xx from the mean position and acts in the opposite direction: F=kxF = -kx.

Displacement in SHM: The position of a particle at any time tt is given by x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi) or x(t)=Asin(ωt+ϕ)x(t) = A \sin(\omega t + \phi), where AA is the amplitude and ω\omega is the angular frequency.

Phase and Phase Constant: The term (ωt+ϕ)(\omega t + \phi) is the phase, representing the state of motion. ϕ\phi is the phase constant or initial phase at t=0t = 0.

Velocity in SHM: Velocity is the rate of change of displacement, given by v(t)=Aωsin(ωt+ϕ)v(t) = -A\omega \sin(\omega t + \phi). In terms of displacement: v=±ωA2x2v = \pm \omega \sqrt{A^2 - x^2}. Velocity is maximum (AωA\omega) at the mean position and zero at the extremes.

Acceleration in SHM: Acceleration is a(t)=ω2Acos(ωt+ϕ)=ω2xa(t) = -\omega^2 A \cos(\omega t + \phi) = -\omega^2 x. Acceleration is always directed towards the mean position and is maximum at the extreme positions.

Energy in SHM: The total mechanical energy is the sum of Kinetic Energy K=12mv2=12k(A2x2)K = \frac{1}{2} m v^2 = \frac{1}{2} k (A^2 - x^2) and Potential Energy U=12kx2U = \frac{1}{2} k x^2. The total energy E=12kA2E = \frac{1}{2} k A^2 remains constant.

Simple Pendulum: For small angular displacements, a simple pendulum executes SHM with a time period T=2πlgT = 2\pi \sqrt{\frac{l}{g}}, where ll is the length and gg is the acceleration due to gravity.

📐Formulae

F=kxF = -kx

a=ω2xa = -\omega^2 x

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

x(t)=Asin(ωt+ϕ)x(t) = A \sin(\omega t + \phi)

v=ωA2x2v = \omega \sqrt{A^2 - x^2}

T=2πmkT = 2\pi \sqrt{\frac{m}{k}} (Spring-Mass System)

T=2πlgT = 2\pi \sqrt{\frac{l}{g}} (Simple Pendulum)

Etotal=12mω2A2=12kA2E_{total} = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2

U=12kx2U = \frac{1}{2} k x^2

K=12k(A2x2)K = \frac{1}{2} k (A^2 - x^2)

💡Examples

Problem 1:

A particle executes SHM with an amplitude of 10 cm10 \text{ cm} and a time period of 4 s4 \text{ s}. Calculate the maximum velocity and maximum acceleration of the particle.

Solution:

Given: A=0.1 mA = 0.1 \text{ m}, T=4 sT = 4 \text{ s}. First, find ω\omega: ω=2πT=2π4=π2 rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s}. Maximum velocity vmax=Aω=0.1×π2=0.05π m/s0.157 m/sv_{max} = A\omega = 0.1 \times \frac{\pi}{2} = 0.05\pi \text{ m/s} \approx 0.157 \text{ m/s}. Maximum acceleration amax=ω2A=(π2)2×0.1=π24×0.10.247 m/s2a_{max} = \omega^2 A = (\frac{\pi}{2})^2 \times 0.1 = \frac{\pi^2}{4} \times 0.1 \approx 0.247 \text{ m/s}^2.

Explanation:

In SHM, maximum velocity occurs at the mean position (x=0x=0) and maximum acceleration occurs at the extreme positions (x=Ax=A).

Problem 2:

At what displacement from the mean position is the kinetic energy of a particle executing SHM equal to three times its potential energy?

Solution:

Condition: K=3UK = 3U. 12k(A2x2)=3(12kx2)\frac{1}{2} k (A^2 - x^2) = 3 \left( \frac{1}{2} k x^2 \right). A2x2=3x2A2=4x2A^2 - x^2 = 3x^2 \Rightarrow A^2 = 4x^2. x2=A24x=±A2x^2 = \frac{A^2}{4} \Rightarrow x = \pm \frac{A}{2}.

Explanation:

By equating the expressions for Kinetic Energy and Potential Energy with the given ratio, we can solve for the specific displacement xx in terms of amplitude AA.

Simple Harmonic Motion (SHM) - Revision Notes & Key Formulas | CBSE Class 11 Physics