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Oscillations and Waves - Simple Harmonic Motion (SHM)

Grade 11ICSEPhysics

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

πŸ”‘Concepts

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Periodic Motion: Motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called the period TT.

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Simple Harmonic Motion (SHM): A specific type of oscillatory motion where the restoring force FF acting on the body is directly proportional to its displacement xx from the mean position and acts in the opposite direction: F=βˆ’kxF = -kx.

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Restoring Force: The force that tends to bring the system back to its equilibrium position. In SHM, F=βˆ’mΟ‰2xF = -m\omega^2 x.

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Amplitude (AA): The maximum displacement of the particle from its mean position on either side.

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Phase and Phase Constant: The state of motion of a particle at any instant is represented by the phase angle (Ο‰t+Ο•)(\omega t + \phi). Ο•\phi is the initial phase at t=0t = 0.

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Energy in SHM: In the absence of friction, the total mechanical energy EE remains constant. It is the sum of Kinetic Energy (KK) and Potential Energy (UU). At the mean position, U=0U=0 and KK is maximum. At extreme positions, K=0K=0 and UU is maximum.

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Simple Pendulum: For small angular displacements, a simple pendulum executes SHM. Its time period depends only on the length ll and acceleration due to gravity gg.

πŸ“Formulae

x(t)=Asin⁑(Ο‰t+Ο•)x(t) = A \sin(\omega t + \phi) or x(t)=Acos⁑(Ο‰t+Ο•)x(t) = A \cos(\omega t + \phi)

Ο‰=2Ο€T=2Ο€f\omega = \frac{2\pi}{T} = 2\pi f

v=dxdt=AΟ‰cos⁑(Ο‰t+Ο•)=Ο‰A2βˆ’x2v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}

a=dvdt=βˆ’Ο‰2Asin⁑(Ο‰t+Ο•)=βˆ’Ο‰2xa = \frac{dv}{dt} = -\omega^2 A \sin(\omega t + \phi) = -\omega^2 x

U=12kx2=12mω2x2U = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2

K=12mv2=12mΟ‰2(A2βˆ’x2)K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)

Etotal=K+U=12mω2A2E_{total} = K + U = \frac{1}{2}m\omega^2 A^2

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

T=2Ο€lgT = 2\pi \sqrt{\frac{l}{g}} (Simple Pendulum)

πŸ’‘Examples

Problem 1:

A body of mass 0.5Β kg0.5\text{ kg} executes SHM with an amplitude of 0.1Β m0.1\text{ m} and a time period of 0.2Β s0.2\text{ s}. Calculate the maximum force acting on the body.

Solution:

  1. Find angular frequency Ο‰\omega: Ο‰=2Ο€T=2Ο€0.2=10π rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\text{ rad/s}.
  2. Find maximum acceleration amaxa_{max}: amax=Ο‰2A=(10Ο€)2Γ—0.1=100Ο€2Γ—0.1=10Ο€2Β m/s2a_{max} = \omega^2 A = (10\pi)^2 \times 0.1 = 100\pi^2 \times 0.1 = 10\pi^2\text{ m/s}^2.
  3. Calculate maximum force FmaxF_{max}: Fmax=mΓ—amax=0.5Γ—10Ο€2=5Ο€2Β NF_{max} = m \times a_{max} = 0.5 \times 10\pi^2 = 5\pi^2\text{ N}. Using Ο€2β‰ˆ9.87\pi^2 \approx 9.87, Fmaxβ‰ˆ49.35Β NF_{max} \approx 49.35\text{ N}.

Explanation:

The maximum force in SHM occurs at the extreme positions where the acceleration is at its maximum value. Using the relationship F=maF = ma and a=βˆ’Ο‰2xa = -\omega^2 x, we derive the peak force.

Problem 2:

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

Solution:

Given K=UK = U. 12mΟ‰2(A2βˆ’x2)=12mΟ‰2x2\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2 A2βˆ’x2=x2A^2 - x^2 = x^2 2x2=A22x^2 = A^2 x2=A22x^2 = \frac{A^2}{2} x=Β±A2x = \pm \frac{A}{\sqrt{2}}

Explanation:

By equating the expressions for Kinetic Energy and Potential Energy, we solve for the displacement xx in terms of the amplitude AA. This occurs at approximately 0.707A0.707A.

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