Physics > Simple Harmonic Motion > 8.0 Force and Energy in Simple Harmonic Motion
Simple Harmonic Motion
1.0 Types of Motion
2.0 Causes of Oscillation
3.0 Solution of the Equation of SHM
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
6.0 Sign Convention of a Simple Harmonic Motion
7.0 How to Write the Simple Harmonic Motion Equation
8.0 Force and Energy in Simple Harmonic Motion
9.0 Basic Differential Equation of SHM
10.0 Method for Calculating Time Period of a Simple Harmonic Motion
10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
11.0 Spring Block System
12.0 Physical Pendulum
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
14.0 Simple Harmonic Oscillation of a Fluid Column
8.3 Total Energy ($E$)
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
Total energy can be obtained by adding potential and kinetic energies Therefore,
$$\begin{equation} \begin{aligned} E = K + U = \frac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) + \frac{1}{2}m{\omega ^2}{x^2} \\ E = \frac{1}{2}m{\omega ^2}{A^2} = \frac{1}{2}k{A^2} \\\end{aligned} \end{equation} $$
Total energy is a constant quantity. It is constant because the force is conservative.
Therefore, we understand that during an oscillation there is continuous exchange of kinetic and potential energies. While moving away from the equilibrium position, the potential energy increases at the expense of the kinetic energy. When the particle moves towards the equilibrium position, the reverse happens.
Figure shows the variation of total energy ($E$), potential energy ($U$) and kinetic energy ($K$) with displacement ($x$).
Example 3. Derive the
(a). Torque ($\tau $),
(b). Kinetic energy ($K$),
(c). Potential energy ($P$) and
(d). Total energy ($E$) for angular SHM.
Solution: (a). Since torque in an angular SHM is proportional to the angular displacement.
Therefore,
$$\begin{equation} \begin{aligned} \tau \propto - \theta \\ \tau = - k\theta \\ I\alpha = - k\theta \\ \alpha = - \frac{k}{I}\theta \\ \alpha = - {\omega ^2}\theta \\ \omega = \sqrt {\frac{k}{I}} \\ T = 2\pi \sqrt {\frac{I}{k}} \\\end{aligned} \end{equation} $$
(b). For a particle undergoing angular SHM, the equation is given by $$\theta = {\theta _o}{\text{sin}}\left( {\omega 't + \phi } \right)$$
where,
$\theta$ = angular displacement at any time $t$.
${\theta _0}$ = angular amplitude
$\omega {\text{'}}$ = angular frequency of SHM
$\phi $ = phase difference
On differentiating, we get the angular velocity ($\omega$),
$$\frac{{d\theta }}{{dt}} = \omega = \frac{{{\theta _o}d{\text{sin}}\left( {\omega 't + \phi } \right)}}{{dt}} = \omega '{\theta _o}{\text{cos}}\left( {\omega 't + \phi } \right)$$
Rotational Kinetic Energy is $$\left( K \right) = \frac{1}{2}I{\omega ^2}$$
where,
$I$ = Moment of inertia about the particle is hinged
$\omega$ = angular velocity
$$\begin{equation} \begin{aligned} K = \frac{1}{2}I{\left\{ {\omega '{\theta _o}{\text{cos}}\left( {\omega 't + \phi } \right)} \right\}^2} \\ K = \frac{1}{2}I{\omega ^{'2}}\theta _o^2co{s^2}\left( {\omega 't + \phi } \right) \\ K = \frac{1}{2}I{\omega ^{'2}}\left\{ {\theta _o^2 - \theta _o^2si{n^2}\left( {\omega 't + \phi } \right)} \right\} \\ K = \frac{1}{2}I{\omega ^{'2}}\left( {\theta _o^2 - {\theta ^2}} \right) \\\end{aligned} \end{equation} $$
(c). Potential energy is $$(P) = \frac{1}{2}k{\theta ^2}$$ As, $k = I{\omega ^{'2}}$ $$P = \frac{1}{2}I{\omega ^{'2}}{\theta ^2}$$
(d). Total Energy is $$\begin{equation} \begin{aligned} \left( E \right) = K + P \\ E = \frac{1}{2}I{\omega ^{'2}}\left( {\theta _o^2 - {\theta ^2}} \right) + \frac{1}{2}I{\omega ^{'2}}{\theta ^2} \\ E = \frac{1}{2}I{\omega ^{'2}}\theta _o^2 \\\end{aligned} \end{equation} $$
