Oscillations • Topic 1 of 3

SHM: Displacement, Velocity & Acceleration

A motion that repeats itself at regular intervals of time is called periodic motion — the hands of a clock, the Earth orbiting the Sun, a vibrating tuning fork. A special kind of periodic motion is oscillatory (or vibratory) motion, in which a body moves to and fro about a fixed mean (equilibrium) position. Every oscillation is periodic, but not every periodic motion is oscillatory (planetary orbits are periodic but not to-and-fro).

The simplest and most important oscillation is Simple Harmonic Motion (SHM). A particle is said to execute SHM when its acceleration is directly proportional to its displacement from the mean position and is always directed towards that mean position. In symbols, $a=-\omega^2 x$, where $x$ is the displacement, $\omega$ is a positive constant called the angular frequency, and the negative sign shows the acceleration (and the restoring force $F=-m\omega^2 x$) points back towards equilibrium. This restoring-force condition, $F\propto -x$, is the defining test for SHM.

  • The displacement of a particle in SHM is $x=A\sin(\omega t+\phi)$, where $A$ is the amplitude (maximum displacement), $(\omega t+\phi)$ is the phase and $\phi$ is the initial phase (epoch) at $t=0$.
  • The velocity is the time-derivative of displacement: $v=\frac{dx}{dt}=A\omega\cos(\omega t+\phi)$. Eliminating $t$ gives the useful relation $v=\omega\sqrt{A^2-x^2}$.
  • The acceleration is $a=\frac{dv}{dt}=-A\omega^2\sin(\omega t+\phi)=-\omega^2 x$, confirming the SHM definition.

Where are these quantities largest? Velocity is maximum at the mean position ($x=0$), where $v_{max}=A\omega$, and zero at the extremes ($x=\pm A$). Acceleration is zero at the mean position and maximum at the extremes, where $a_{max}=\omega^2 A$. Displacement and acceleration are always opposite in sign (out of phase by $\pi$), while velocity is ahead of displacement by a phase of $\frac{\pi}{2}$.

Time period and frequency. The time period $T$ is the time for one complete oscillation, and the frequency $f$ is the number of oscillations per second. They are linked to the angular frequency by $\omega=\frac{2\pi}{T}=2\pi f$, so $T=\frac{2\pi}{\omega}$ and $f=\frac{1}{T}$. The SI unit of frequency is the hertz (Hz). A neat geometric picture: SHM is the projection of uniform circular motion onto a diameter — a reference particle moving in a circle of radius $A$ at angular speed $\omega$ casts a shadow that performs SHM, which is why the equations involve $\sin$ and $\cos$.

Displacement-time graph of SHMtx+A-AT (one period)
Solved Examples
1
Worked Example
A particle executes SHM described by $x=5\sin\left(4\pi t\right)$ (in cm, with $t$ in seconds). Find the amplitude, angular frequency, time period and frequency.
Solution
  1. Step 1: Compare with $x=A\sin(\omega t+\phi)$. Here $A=5\ \text{cm}$ and $\omega=4\pi\ \text{rad/s}$, with $\phi=0$.
  2. Step 2: Time period $T=\frac{2\pi}{\omega}=\frac{2\pi}{4\pi}=0.5\ \text{s}$.
  3. Step 3: Frequency $f=\frac{1}{T}=\frac{1}{0.5}=2\ \text{Hz}$.

Answer: $A=5\ \text{cm}$, $\omega=4\pi\ \text{rad/s}$, $T=0.5\ \text{s}$, $f=2\ \text{Hz}$.

2
Worked Example
A body in SHM has amplitude 0.1 m and angular frequency $5\ \text{rad/s}$. Find its maximum velocity and maximum acceleration.
Solution
  1. Step 1: Maximum velocity occurs at the mean position: $v_{max}=A\omega$.
  2. Step 2: $v_{max}=0.1\times5=0.5\ \text{m/s}$.
  3. Step 3: Maximum acceleration occurs at the extremes: $a_{max}=A\omega^2=0.1\times5^2=2.5\ \text{m/s}^2$.

Answer: $v_{max}=0.5\ \text{m/s}$, $a_{max}=2.5\ \text{m/s}^2$.

3
Worked Example
A particle performs SHM with amplitude 8 cm and period 4 s. Find its velocity when the displacement is 4 cm.
Solution
  1. Step 1: $\omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{rad/s}$.
  2. Step 2: Use $v=\omega\sqrt{A^2-x^2}=\frac{\pi}{2}\sqrt{8^2-4^2}=\frac{\pi}{2}\sqrt{48}$.
  3. Step 3: $\sqrt{48}=6.93\ \text{cm}$, so $v=\frac{\pi}{2}\times6.93\approx10.9\ \text{cm/s}$.

Answer: $v\approx10.9\ \text{cm/s}\ (\approx0.109\ \text{m/s})$.

4
Worked Example
The acceleration of a particle in SHM is $-16\ \text{m/s}^2$ when its displacement is $0.04\ \text{m}$. Find the time period of the motion.
Solution
  1. Step 1: For SHM, $a=-\omega^2 x$, so $\omega^2=\frac{|a|}{|x|}=\frac{16}{0.04}=400$.
  2. Step 2: $\omega=\sqrt{400}=20\ \text{rad/s}$.
  3. Step 3: $T=\frac{2\pi}{\omega}=\frac{2\pi}{20}=\frac{\pi}{10}\approx0.314\ \text{s}$.

Answer: $T\approx0.314\ \text{s}$.

5
Worked Example
A particle starts from the mean position and executes SHM with period 8 s and amplitude 6 cm. Find its displacement after 1 s.
Solution
  1. Step 1: Starting from the mean position, $x=A\sin(\omega t)$ with $\omega=\frac{2\pi}{T}=\frac{2\pi}{8}=\frac{\pi}{4}\ \text{rad/s}$.
  2. Step 2: At $t=1\ \text{s}$, $x=6\sin\left(\frac{\pi}{4}\times1\right)=6\sin\left(\frac{\pi}{4}\right)$.
  3. Step 3: $\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}=0.707$, so $x=6\times0.707\approx4.24\ \text{cm}$.

Answer: $x\approx4.24\ \text{cm}$.

6
Worked Example
A particle in SHM has a maximum velocity of $0.4\ \text{m/s}$ and a maximum acceleration of $0.8\ \text{m/s}^2$. Find the amplitude and the time period.
Solution
  1. Step 1: $v_{max}=A\omega=0.4$ and $a_{max}=A\omega^2=0.8$.
  2. Step 2: Divide: $\frac{a_{max}}{v_{max}}=\frac{A\omega^2}{A\omega}=\omega=\frac{0.8}{0.4}=2\ \text{rad/s}$.
  3. Step 3: Then $A=\frac{v_{max}}{\omega}=\frac{0.4}{2}=0.2\ \text{m}$ and $T=\frac{2\pi}{\omega}=\frac{2\pi}{2}=\pi\approx3.14\ \text{s}$.

Answer: $A=0.2\ \text{m}$, $T=\pi\approx3.14\ \text{s}$.

Key Points

  • Periodic motion repeats at fixed intervals; oscillatory motion is to-and-fro about a mean position — all oscillations are periodic, not vice versa.
  • SHM is defined by $a=-\omega^2 x$ (restoring acceleration $\propto$ displacement, directed to the mean position); equivalently $F=-m\omega^2 x$.
  • Displacement $x=A\sin(\omega t+\phi)$, velocity $v=A\omega\cos(\omega t+\phi)=\omega\sqrt{A^2-x^2}$, acceleration $a=-\omega^2 x$.
  • Velocity is maximum ($A\omega$) at the mean position and zero at the extremes; acceleration is zero at the mean and maximum ($\omega^2 A$) at the extremes.
  • Angular frequency $\omega=\frac{2\pi}{T}=2\pi f$; SHM is the projection of uniform circular motion on a diameter.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A motion is simple harmonic if the acceleration of the body is:
Explanation: SHM requires $a=-\omega^2 x$ — acceleration proportional to displacement and always directed towards the mean position.
Q2.In SHM, the velocity of the particle is maximum at the:
Explanation: Since $v=\omega\sqrt{A^2-x^2}$, $v$ is greatest when $x=0$, i.e. at the mean position, where $v_{max}=A\omega$.
Q3.For a particle in SHM, the relation between angular frequency $\omega$ and time period $T$ is:
Explanation: By definition $\omega=\frac{2\pi}{T}=2\pi f$.
Q4.The acceleration of a particle executing SHM is zero when the displacement is:
Explanation: Since $a=-\omega^2 x$, acceleration is zero only when $x=0$, i.e. at the mean position.
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