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Energy in SHM


There are two types of energy involved in simple harmonic motion: kinetic energy $K$ and potential energy $U$ given by \begin{align} U & =\frac{1}{2} kx^2, \\ K & = \frac{1}{2} mv^2. \end{align}

At the equilibrium position, all of its energy is in the form of potential energy, which is stored in the system due to the deformation of the spring or other restoring force. As the object is displaced from its equilibrium position, the potential energy is converted into kinetic energy. At the maximum displacement, the kinetic energy is zero and the potential energy is at a maximum.


The relationship between $K$ and $U$ in SHM is often represented graphically by an energy diagram, which shows how the energy of the system varies with the position of the object. This diagram can be used to calculate various parameters of the system, such as the amplitude and frequency of the motion, and to analyze the behavior of the system under different conditions.

The total mechanical energy ($E$) of the system, which is the sum of the kinetic energy and potential energy, remains constant throughout the motion of the object. \begin{align} E=U+K=\frac{1}{2}m\omega^2 A^2. \end{align}

The potential and kinetic energy in SHM oscillate with time. The time period of $U$ (or $K$) oscillation is half of the SHM time period. These energies varies with frequency twice of SHM frequency. This can be shown by subtituting $x=A\sin\omega t$ and $v=A\omega\cos\omega t$ in the expressions of $U$ and $K$.


Problems from IIT JEE

Problem (IIT JEE 1987): A particle executes SHM with a frequency $f$. The frequency with which its kinetic energy oscillates is_______.

Solution: The displacement of a particle undergoing SHM varies with time $t$ as \begin{align} x=A\sin2\pi f t, \end{align} where $A$ is the amplitude and $f$ is the frequency. The velocity of the particle is \begin{align} v=\mathrm{d}x/\mathrm{d}t=2\pi Af\cos2\pi f t. \end{align} The kinetic energy of a particle of mass $m$ moving with velocity $v$ is \begin{align} K&=\frac{1}{2}mv^2 \\ &=\frac{1}{2} m (2\pi f)^2 A^2\cos^2(2\pi f t)\nonumber \\ &=m\pi^2 f^2 A^2(1+\cos4\pi f t).\nonumber \end{align} Thus, the kinetic energy varies with a frequency $2f$.

Problem (IIT JEE 1989): A linear harmonic oscillator of force constant $2\times{10}^{6}$ N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its

  1. maximum potential energy is 100 J.
  2. maximum kinetic energy is 100 J.
  3. maximum potential energy is 160 J.
  4. maximum potential energy is zero.

Solution: The potential energy $U(x)$, kinetic energy $K(x)$, and total energy $E$ of a linear harmonic oscillator are related by \begin{align} & U(x)+K(x)=E. \end{align}


The potential energy of a linear harmonic oscillator of force constant $k$ at a displacement $x$ is given by \begin{align} U(x)=U_0+\tfrac{1}{2}kx^2, \end{align} where $U_0$ is a constant. The kinetic energy of the oscillator becomes zero at the maximum displacement ($x=\pm A$). Substitute $K(x)=0$, $x=A=0.01$ m, and $k=2\times{10}^{6}$ N/m in above equations and simplify to get \begin{align} U_0&=E-\frac{1}{2}kA^2\nonumber\\ &=160-\frac{1}{2}(2\times{10}^{6})(0.01)^2\\ &=60\;\mathrm{J}. \end{align} Use above equations to get the minimum and maximum potential and kinetic energies as \begin{align} x=0:& && U_\text{min}=60\,\mathrm{J}, && K_\text{max}=100\,\mathrm{J};\nonumber\\ x=\pm A:& && U_\text{max}=160\,\mathrm{J}, && K_\text{min}=0.\nonumber \end{align}


  1. Simple Harmonic Motion
JEE Physics Solved Problems in Mechanics