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Oscillations - Energy in SHM

Grade 11CBSEPhysics

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

πŸ”‘Concepts

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The total mechanical energy (EE) of a particle performing Simple Harmonic Motion (SHM) is the sum of its Kinetic Energy (KK) and Potential Energy (UU).

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Potential Energy (UU) is due to the displacement of the particle from its mean position. It is given by U=12kx2U = \frac{1}{2} k x^2, where kk is the force constant and xx is the displacement.

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Kinetic Energy (KK) is due to the velocity of the particle. It is maximum at the mean position (x=0x = 0) and zero at the extreme positions (x=Β±Ax = \pm A).

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In an ideal SHM (no damping), the Total Mechanical Energy remains constant at all points during the oscillation, satisfying the Law of Conservation of Energy.

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The Potential Energy and Kinetic Energy vary periodically. While the displacement varies with frequency ff, the energies KK and UU vary with a frequency of 2f2f.

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At the mean position (x=0x=0), E=KmaxE = K_{max}. At the extreme positions (x=Ax=A), E=UmaxE = U_{max}.

πŸ“Formulae

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

K=12mv2=12mΟ‰2(A2βˆ’x2)K = \frac{1}{2} m v^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

v=Ο‰A2βˆ’x2v = \omega \sqrt{A^2 - x^2}

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

Uavg=Kavg=14mω2A2=12EtotalU_{avg} = K_{avg} = \frac{1}{4} m \omega^2 A^2 = \frac{1}{2} E_{total}

πŸ’‘Examples

Problem 1:

A particle of mass 0.2Β kg0.2 \text{ kg} executes SHM with an amplitude of 0.05Β m0.05 \text{ m} and an angular frequency Ο‰=10Β rad/s\omega = 10 \text{ rad/s}. Calculate the total energy and the kinetic energy when the displacement is 0.03Β m0.03 \text{ m}.

Solution:

  1. Total Energy: E=12mω2A2=12(0.2)(10)2(0.05)2=0.1×100×0.0025=0.025 JE = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} (0.2) (10)^2 (0.05)^2 = 0.1 \times 100 \times 0.0025 = 0.025 \text{ J}.
  2. Kinetic Energy at x=0.03Β mx = 0.03 \text{ m}: K=12mΟ‰2(A2βˆ’x2)=12(0.2)(10)2(0.052βˆ’0.032)=10Γ—(0.0025βˆ’0.0009)=10Γ—0.0016=0.016Β JK = \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} (0.2) (10)^2 (0.05^2 - 0.03^2) = 10 \times (0.0025 - 0.0009) = 10 \times 0.0016 = 0.016 \text{ J}.

Explanation:

The total energy is calculated using the maximum amplitude, while the kinetic energy at a specific point is found by subtracting the potential energy at that displacement from the total energy.

Problem 2:

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

Solution:

Given K=UK = U. 12k(A2βˆ’x2)=12kx2\frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k 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 KK and UU, we find that the energies are equal when the displacement is 12\frac{1}{\sqrt{2}} times the amplitude.

Energy in SHM - Revision Notes & Key Formulas | CBSE Class 11 Physics