SHM Equations — Revision Notes

⚡ 30-Second Revision

Displacement: — $x(t) = A \sin(\omega t + \phi)$ or $A \cos(\omega t + \phi)$
Velocity: — $v(t) = A\omega \cos(\omega t + \phi)$ or $-A\omega \sin(\omega t + \phi)$
Maximum Velocity: — $v_{max} = A\omega$
Acceleration: — $a(t) = -A\omega^2 \sin(\omega t + \phi)$ or $-A\omega^2 \cos(\omega t + \phi)$
Acceleration in terms of x: — $a(t) = -\omega^2 x(t)$
Maximum Acceleration: — $a_{max} = A\omega^2$
Angular Frequency: — $\omega = \sqrt{k/m}$ (spring-mass), $\omega = \sqrt{g/L}$ (pendulum)
Time Period: — $T = 2\pi/\omega$
Frequency: — $f = 1/T = \omega/(2\pi)$
Kinetic Energy: — $KE = \frac{1}{2}mv^2$
Potential Energy: — $PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$
Total Energy: — $E = \frac{1}{2}m A^2\omega^2 = \frac{1}{2}kA^2$
Velocity-Displacement Relation: — $v = \pm \omega \sqrt{A^2 - x^2}$

2-Minute Revision

SHM equations describe the position, velocity, and acceleration of an object undergoing Simple Harmonic Motion. The core idea is that the restoring force is proportional to displacement ($F = -kx$), leading to the differential equation $\frac{d^2x}{dt^2} + \omega^2x = 0$.

The solutions are sinusoidal: $x(t) = A \sin(\omega t + \phi)$ for displacement, where $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is initial phase. Velocity is the time derivative of displacement, $v(t) = A\omega \cos(\omega t + \phi)$, with maximum value $A\omega$ at equilibrium.

Acceleration is the time derivative of velocity, $a(t) = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x(t)$, with maximum value $A\omega^2$ at the extreme positions. Remember that velocity leads displacement by $\pi/2$ and acceleration leads velocity by $\pi/2$.

The total mechanical energy ($E = \frac{1}{2}kA^2 = \frac{1}{2}m A^2\omega^2$) remains constant, continuously converting between kinetic and potential forms. Key relations are $\omega = 2\pi f = 2\pi/T$.

For a mass-spring system, $\omega = \sqrt{k/m}$, and for a simple pendulum (small angles), $\omega = \sqrt{g/L}$.

5-Minute Revision

Simple Harmonic Motion (SHM) is a fundamental oscillatory motion characterized by a restoring force directly proportional to displacement and directed towards equilibrium ($F = -kx$). This leads to the defining differential equation $\frac{d^2x}{dt^2} + \omega^2x = 0$, where $\omega = \sqrt{k/m}$ for a spring-mass system. The solutions to this equation give us the SHM equations:

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Displacement: — $x(t) = A \sin(\omega t + \phi)$ (or $A \cos(\omega t + \phi)$). $A$ is the maximum displacement (amplitude), $\omega$ is the angular frequency (rad/s), and $\phi$ is the initial phase (rad), determining the starting point at $t=0$.

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Velocity: — $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum velocity is $v_{max} = A\omega$, occurring at the equilibrium position ($x=0$). Velocity is zero at the extreme positions ($x=\pm A$). Velocity leads displacement by a phase of $\pi/2$.

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Acceleration: — $a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x(t)$. The maximum acceleration is $a_{max} = A\omega^2$, occurring at the extreme positions ($x=\pm A$). Acceleration is zero at the equilibrium position ($x=0$). Acceleration leads velocity by $\pi/2$ and is $\pi$ (or $180^\circ$) out of phase with displacement.

Related Parameters:

Time Period (T): — $T = \frac{2\pi}{\omega}$.
Frequency (f): — $f = \frac{1}{T} = \frac{\omega}{2\pi}$.

Energy in SHM:

Kinetic Energy (KE): — $KE = \frac{1}{2}mv^2$. Maximum at equilibrium, zero at extremes.
Potential Energy (PE): — $PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$. Maximum at extremes, zero at equilibrium.
Total Mechanical Energy (E): — $E = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m A^2\omega^2$. This is constant throughout the motion.

Key Relationship: $v = \pm \omega \sqrt{A^2 - x^2}$. This formula is extremely useful for finding velocity at any displacement or for determining amplitude from given $x$ and $v$.

Example: A particle has SHM described by $x(t) = 0.1 \cos(4t)$. Find its maximum velocity and acceleration. Here, $A = 0.1\text{ m}$ and $\omega = 4\text{ rad/s}$. $v_{max} = A\omega = 0.1 \times 4 = 0.4\text{ m/s}$. $a_{max} = A\omega^2 = 0.1 \times (4)^2 = 0.1 \times 16 = 1.6\text{ m/s}^2$.

Prelims Revision Notes

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SHM Definition: — Restoring force $F \propto -x$, so $F = -kx$. This leads to $a = -\omega^2 x$.
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Displacement Equation: — $x(t) = A \sin(\omega t + \phi)$ or $x(t) = A \cos(\omega t + \phi)$.

* $A$: Amplitude (max displacement). * $\omega$: Angular frequency (rad/s). $\omega = 2\pi f = 2\pi/T$. * $\phi$: Initial phase (rad). Determines $x(0)$.

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Velocity Equation: — $v(t) = \frac{dx}{dt}$.

* If $x(t) = A \sin(\omega t + \phi)$, then $v(t) = A\omega \cos(\omega t + \phi)$. * If $x(t) = A \cos(\omega t + \phi)$, then $v(t) = -A\omega \sin(\omega t + \phi)$. * Maximum Velocity: $v_{max} = A\omega$. Occurs at $x=0$ (equilibrium).

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Acceleration Equation: — $a(t) = \frac{dv}{dt}$.

* If $x(t) = A \sin(\omega t + \phi)$, then $a(t) = -A\omega^2 \sin(\omega t + \phi)$. * If $x(t) = A \cos(\omega t + \phi)$, then $a(t) = -A\omega^2 \cos(\omega t + \phi)$. * Relation to displacement: $a(t) = -\omega^2 x(t)$. * Maximum Acceleration: $a_{max} = A\omega^2$. Occurs at $x=\pm A$ (extreme positions).

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Phase Relationships:

* Velocity leads displacement by $\pi/2$ (or $90^\circ$). * Acceleration leads velocity by $\pi/2$ (or $90^\circ$). * Acceleration is $180^\circ$ (or $\pi$) out of phase with displacement.

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Time Period (T) & Frequency (f):

* $T = 2\pi/\omega$ * $f = 1/T = \omega/(2\pi)$ * For spring-mass: $\omega = \sqrt{k/m}$, $T = 2\pi\sqrt{m/k}$. * For simple pendulum (small angles): $\omega = \sqrt{g/L}$, $T = 2\pi\sqrt{L/g}$.

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Energy in SHM:

* Kinetic Energy: $KE = \frac{1}{2}mv^2$. Max at equilibrium, zero at extremes. * Potential Energy: $PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$. Max at extremes, zero at equilibrium. * Total Energy: $E = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m A^2\omega^2$. Constant.

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Velocity-Displacement Relation: — $v = \pm \omega \sqrt{A^2 - x^2}$. Use this to find $v$ at any $x$, or $A$ given $x, v, \omega$.
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Initial Conditions: — Use $x(0)$ and $v(0)$ to find $A$ and $\phi$. $A = \sqrt{x(0)^2 + (v(0)/\omega)^2}$. $\tan\phi = -v(0)/(\omega x(0))$ (if using $A \cos(\omega t + \phi)$).

Vyyuha Quick Recall

A-V-A: Amplitude, Velocity, Acceleration. Remember the phase shifts: 'V' leads 'X' by 90, 'A' leads 'V' by 90. So 'A' is 180 opposite 'X'. For formulas: 'A' has no $\omega$, 'V' has one $\omega$, 'A' (acceleration) has two $\omega$'s (squared!).