Energy in Simple Harmonic Motion (SHM)

Simple Harmonic Motion (SHM) is a fundamental concept in physics, describing oscillatory motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. Understanding the energy transformations within SHM is crucial for solving problems in competitive exams like JEE.

Types of Energy in SHM

In SHM, energy oscillates between kinetic energy (energy of motion) and potential energy (stored energy due to position or configuration). The total mechanical energy of the system remains constant, assuming no dissipative forces like friction.

Kinetic Energy (KE)

Kinetic energy is the energy possessed by an object due to its motion. In SHM, the velocity of the oscillating object varies with its position. The formula for kinetic energy is given by $KE = \frac{1}{2}mv^2$ , where $m$ is the mass and $v$ is the velocity.

Velocity in SHM.

The velocity in SHM is maximum at the mean position and zero at the extreme positions. It can be expressed as $v = \pm \omega \sqrt{A^2 - x^2}$ , where $\omega$ is the angular frequency, $A$ is the amplitude, and $x$ is the displacement from the mean position.

The velocity of an object undergoing SHM can be derived from its displacement equation, $x(t) = A \cos(\omega t + \phi)$ . Differentiating with respect to time gives the velocity: $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$ . The maximum velocity occurs when $\sin(\omega t + \phi) = \pm 1$ , so $v_{max} = A\omega$ . Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ , we can relate velocity to displacement: $v^2 = A^2\omega^2 \sin^2(\omega t + \phi) = A^2\omega^2 (1 - \cos^2(\omega t + \phi)) = \omega^2 (A^2 - A^2 \cos^2(\omega t + \phi)) = \omega^2 (A^2 - x^2)$ . Therefore, $v = \pm \omega \sqrt{A^2 - x^2}$ .

Substituting this into the KE formula: $KE = \frac{1}{2}m (\omega^2 (A^2 - x^2)) = \frac{1}{2}m\omega^2 (A^2 - x^2)$ .

At what position in SHM is the kinetic energy maximum?

At the mean position (equilibrium position, where x=0).

At what position in SHM is the kinetic energy minimum (zero)?

At the extreme positions (maximum displacement, where x = ±A).

Potential Energy (PE)

Potential energy in SHM is typically associated with the restoring force. For a mass-spring system, the restoring force is $F = -kx$ , where $k$ is the spring constant. The potential energy stored in the spring is given by $PE = \frac{1}{2}kx^2$ .

Relationship between force and potential energy.

Potential energy is the work done against the restoring force. For a linear restoring force $F = -kx$ , the potential energy is $PE = \int F dx = \int (-kx) dx = -\frac{1}{2}kx^2 + C$ . If we set PE=0 at x=0, then $C=0$ , giving $PE = \frac{1}{2}kx^2$ .

The work done by the restoring force as the object moves from position $x_1$ to $x_2$ is $W = \int_{x_1}^{x_2} F dx$ . The change in potential energy is $\Delta PE = -W = -\int_{x_1}^{x_2} F dx$ . For SHM, $F = -kx$ . So, $\Delta PE = -\int_{x_1}^{x_2} (-kx) dx = \int_{x_1}^{x_2} kx dx = \frac{1}{2}kx^2\Big|_{x_1}^{x_2} = \frac{1}{2}k(x_2^2 - x_1^2)$ . If we define the potential energy to be zero at the equilibrium position ( $x=0$ ), then $PE(x) = \frac{1}{2}kx^2$ .

We know that $k = m\omega^2$ . Substituting this, we get $PE = \frac{1}{2}m\omega^2 x^2$ .

At what position in SHM is the potential energy maximum?

At the extreme positions (maximum displacement, where x = ±A).

At what position in SHM is the potential energy minimum (zero)?

At the mean position (equilibrium position, where x=0).

Total Mechanical Energy (E)

The total mechanical energy in SHM is the sum of kinetic energy and potential energy: $E = KE + PE$ . Since energy is conserved (in the absence of damping), the total energy remains constant throughout the motion.

Total energy is constant and depends on amplitude.

The total energy can be calculated at any point. At the extreme positions (x = ±A), velocity is zero, so $KE = 0$ . Thus, $E = PE_{max} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$ . At the mean position (x = 0), potential energy is zero, so $E = KE_{max} = \frac{1}{2}mv_{max}^2 = \frac{1}{2}m(A\omega)^2 = \frac{1}{2}m\omega^2 A^2$ .

The total mechanical energy $E$ is the sum of kinetic and potential energy: $E = KE + PE = \frac{1}{2}m\omega^2 (A^2 - x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2$ . This shows that the total energy is constant and depends only on the mass ( $m$ ), angular frequency ( $\omega$ ), and amplitude ( $A$ ) of the oscillation. Since $k = m\omega^2$ , the total energy can also be written as $E = \frac{1}{2}kA^2$ .

This diagram illustrates the interplay between kinetic and potential energy during one cycle of SHM. At the mean position (x=0), all energy is kinetic. As the object moves towards an extreme position, kinetic energy decreases, and potential energy increases. At the extreme positions (x=±A), all energy is potential, and kinetic energy is zero. The total energy remains constant throughout.

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Text-based content

Library pages focus on text content

Position	Displacement (x)	Kinetic Energy (KE)	Potential Energy (PE)	Total Energy (E)
Mean Position	0	Maximum ( $\frac{1}{2}m\omega^2 A^2$ )	Minimum (0)	Constant ( $\frac{1}{2}m\omega^2 A^2$ )
Extreme Positions	±A	Minimum (0)	Maximum ( $\frac{1}{2}m\omega^2 A^2$ )	Constant ( $\frac{1}{2}m\omega^2 A^2$ )
Intermediate Position	x	$\frac{1}{2}m\omega^2 (A^2 - x^2)$	$\frac{1}{2}m\omega^2 x^2$	Constant ( $\frac{1}{2}m\omega^2 A^2$ )

Key takeaway: In SHM, energy continuously transforms between kinetic and potential forms, but the total mechanical energy remains constant, directly proportional to the square of the amplitude.

Energy in Different SHM Systems

The principles of energy in SHM apply to various physical systems, including mass-spring systems, simple pendulums (for small oscillations), and LC circuits (in terms of energy stored in the inductor and capacitor).

For a simple pendulum with mass $m$ , length $L$ , and small angle of oscillation $\theta$ , the restoring force is approximately $-mg\theta$ . The potential energy is $PE \approx \frac{1}{2}mgL\theta^2$ , and the kinetic energy is $KE = \frac{1}{2}mv^2$ . The angular frequency is $\omega = \sqrt{g/L}$ . The total energy is $E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m(g/L)A^2$ , where $A$ is the maximum angular displacement.

Learning Resources

Simple Harmonic Motion - Physics Classroom(documentation)

Provides a clear explanation of kinetic, potential, and total energy in SHM with helpful diagrams and examples.

Energy in SHM - Khan Academy(video)

A video tutorial explaining the concepts of kinetic, potential, and total energy in SHM, including derivations.

Simple Harmonic Motion - Wikipedia(wikipedia)

Offers a comprehensive overview of SHM, with a dedicated section on energy conservation and transformations.

JEE Physics: Simple Harmonic Motion - Energy(video)

A YouTube video specifically tailored for JEE preparation, focusing on energy aspects of SHM with problem-solving.

Understanding Energy in SHM - Physics Stack Exchange(blog)

A forum discussion with expert explanations and clarifications on common doubts regarding energy in SHM.

Gravitation and SHM - IIT JEE Study Material(documentation)

Comprehensive study material for IIT JEE, covering SHM and its relation to gravitation, including energy concepts.

Energy Conservation in Oscillations - MIT OpenCourseware(video)

Lecture segment from MIT OpenCourseware focusing on energy conservation in oscillatory systems, applicable to SHM.

SHM: Energy and Power - Byju's(documentation)

Explains SHM, including energy considerations, with examples and practice questions relevant to competitive exams.

The Physics of Simple Harmonic Motion - HyperPhysics(documentation)

A detailed resource covering various aspects of SHM, including energy, with concise explanations and formulas.

JEE Main Physics: SHM - Energy Problems(video)

A problem-solving session on energy in SHM, specifically designed for JEE Main aspirants.