See Our New JEE Book on Amazon

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium position, and the restoring force acting on the object is proportional to its displacement from the equilibrium position. A classic example of simple harmonic motion is the motion of a mass attached to a spring.

Consider a spring of spring constant $k$ and natural length $l$. One end of the spring is fixed and another end is attached to a block of mass $m$. The equilibrium position of the mass is at $x=0$ (at a distance $l+mg/k$ from the fixed end). This spring-mass system executes SHM when the mass is slightly displaced and released from its equilibrium position.

In SHM, the acceleration is proportional to the displacement $x$ and its direction is towards the equilibrium position, \begin{align} a&=\frac{\mathrm{d}^2x}{\mathrm{d}t^2} =-\frac{k}{m}x =-\omega^2 x \end{align}

The **angular frquency** of oscillation is related to the spring constant $k$ and mass $m$ by
\begin{align}
\omega=\sqrt{\frac{k}{m}}.
\end{align}
The frequency $\nu$ is the number of oscillations executed by the system in one second. It is related to the angular frequency by
\begin{align}
\nu=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}.
\end{align}
The frequency increases with decrease in mass.

The **time period** of oscillation is the time between two consecutive passages of the system in the same direction through the same position. It is given by
\begin{align}
T=\frac{1}{\nu}=2\pi\sqrt{\frac{m}{k}}
\end{align}

The **displacement** varies with time $t$ as
\begin{align}
x=A\sin(\omega t+\phi),
\end{align}
where $\phi$ is the initial phase, a constant. The amplitude $A$ is the maximum displacement of the system.

The **velocity** is given by
\begin{align}
v=A\omega\cos(\omega t+\phi)=\pm \omega\sqrt{A^2-x^2}
\end{align}

The simple pendulum, physical pendulum and torsional pendulum are other examples of SHM.

**Problem (IIT JEE 2011):**
A point mass is subjected to two simultaneous sinusoidal displacements in $x$ direction, $x_{1}(t)=A\sin(\omega t)$ and $x_2(t)=A\sin\left(\omega t+\frac{2\pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B\sin(\omega t+\phi)$ brings the mass to complete rest. The value of $B$ and $\phi$ are,

- $\sqrt{2} A$, $3\pi/4$
- $A$, $4\pi/3$
- $\sqrt{3} A$, $5\pi/6$
- $A$, $\pi/3$

**Solution:**
Adding $x_3(t)$ brings the mass to complete rest. Thus, $x_1(t)+x_2(t)+x_3(t)=0$, which gives,
\begin{align}
B\sin(\omega t+\phi)&=-A \left[\sin (\omega t)+\sin \left(\omega t+2\pi/3\right) \right]\nonumber\\
&=-A \left[2\sin\left(\omega t+\pi/3\right) \cos(-\pi/3)\right]\nonumber\\
&=-A\sin\left(\omega t+\pi/3\right) \\
&=A\sin\left(\omega t+4\pi/3\right)\nonumber
\end{align}

*Aliter:* The problem can be solved graphically using vector addition. The displacement $x_1(t)$ can be represented by a vector $\vec{v}_1=A\angle 0$ and displacement $x_2(t)$ by $\vec{v}_2=A\angle (2\pi/3)$ (see figure). The resultant of these two is $\vec{v}=A\angle(\pi/3)$. The resultant of the three is zero if displacement $x_3(t)$ is $-\vec{v}=A\angle(4\pi/3)$.