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Oscillations and Waves - Energy in SHM

Grade 11ICSEPhysics

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

🔑Concepts

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The Total Mechanical Energy (EE) in Simple Harmonic Motion (SHM) is the sum of its Kinetic Energy (KK) and Potential Energy (UU). In an ideal system without friction, this total energy remains constant: E=K+UE = K + U.

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Potential Energy (UU): This is the energy stored due to the displacement from the mean position. It is given by U=12kx2U = \frac{1}{2}kx^2, where kk is the force constant (k=mω2k = m\omega^2). It is zero at the mean position (x=0x = 0) and maximum at the extreme positions (x=±Ax = \pm A).

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Kinetic Energy (KK): This is the energy due to the velocity of the particle. It is given by K=12mv2K = \frac{1}{2}mv^2. Since v=ωA2−x2v = \omega\sqrt{A^2 - x^2}, the kinetic energy is K=12mω2(A2−x2)K = \frac{1}{2}m\omega^2(A^2 - x^2). It is maximum at the mean position and zero at the extreme positions.

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Total Energy (EE): By adding UU and KK, we get E=12mω2A2E = \frac{1}{2}m\omega^2A^2. This shows that the total energy is independent of the displacement xx and depends only on the mass (mm), angular frequency (ω\omega), and amplitude (AA).

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Energy-Displacement Graphs: The graph of UU vs xx is a parabola opening upwards, and KK vs xx is a parabola opening downwards. The Total Energy EE is represented by a horizontal line constant across all xx.

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Frequency of Energy: While the displacement and velocity of SHM have a frequency ff, the potential and kinetic energies oscillate with a frequency of 2f2f because they reach their maximum twice in one complete cycle of oscillation.

📐Formulae

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

K=12mω2(A2−x2)K = \frac{1}{2} m \omega^2 (A^2 - x^2)

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

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

v=±ωA2−x2v = \pm \omega \sqrt{A^2 - x^2}

💡Examples

Problem 1:

A particle of mass 0.2 kg0.2 \text{ kg} executes SHM with an amplitude of 0.05 m0.05 \text{ m} and a time period of 0.4π s0.4 \pi \text{ s}. Calculate (i) the total energy and (ii) the kinetic energy at a distance of 0.03 m0.03 \text{ m} from the mean position.

Solution:

  1. Find angular frequency: ω=2πT=2π0.4π=5 rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4\pi} = 5 \text{ rad/s}.
  2. Total Energy (EE): E=12mω2A2=12×0.2×(5)2×(0.05)2=0.1×25×0.0025=0.00625 JE = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} \times 0.2 \times (5)^2 \times (0.05)^2 = 0.1 \times 25 \times 0.0025 = 0.00625 \text{ J}.
  3. Kinetic Energy (KK) at x=0.03 mx = 0.03 \text{ m}: K=12mω2(A2−x2)=12×0.2×52×(0.052−0.032)=2.5×(0.0025−0.0009)=2.5×0.0016=0.004 JK = \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} \times 0.2 \times 5^2 \times (0.05^2 - 0.03^2) = 2.5 \times (0.0025 - 0.0009) = 2.5 \times 0.0016 = 0.004 \text{ J}.

Explanation:

The total energy is calculated using the maximum displacement (amplitude), while kinetic energy is calculated by subtracting the potential energy at that specific point 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. 12mω2(A2−x2)=12mω2x2\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2. Canceling common terms: A2−x2=x2A^2 - x^2 = x^2. A2=2x2  ⟹  x2=A22A^2 = 2x^2 \implies x^2 = \frac{A^2}{2}. x=±A2x = \pm \frac{A}{\sqrt{2}}.

Explanation:

By setting the equations for KK and UU equal to each other, we solve for xx in terms of the amplitude AA.

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