Simple harmonic motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement from a fixed equilibrium position and is directed towards that position. It forms the foundation for understanding oscillatory phenomena, wave propagation, and acoustics in classical mechanics.
Kinematics of SHM, displacement, velocity, acceleration equations, time period and frequency of mass-spring systems, simple pendulum, and energy conservation.
Phase and phase difference, composition of two SHM lines (orthogonal and parallel), angular harmonic oscillators, physical pendulum, torsional pendulum, and differential formulation.
Conditions: Horizontal frictionless surface, ideal spring obeying Hooke's law
Result: $T = 2\pi \sqrt{\frac{m}{k}}$
Conditions: Small amplitude approximation (θ << 1 radian), inextensible string
Result: $T = 2\pi \sqrt{\frac{l}{g}}$
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Displacement in SHM | $$x(t) = A \cos(\omega t + \phi)$$ | A = Amplitude, $\omega = Angular frequency$, t = time, $\phi = Phase constant$ m | Standard initial conditions |
| Velocity in SHM | $$v(t) = -A \omega \sin(\omega t + \phi) = \pm \omega \sqrt{A^2 - x^2}$$ | v = velocity, x = displacement m s^-1 | Valid for any conservative SHM |
| Acceleration in SHM | $$a(t) = -A \omega^2 \cos(\omega t + \phi) = -\omega^2 x$$ | a = acceleration m s^-2 | Always directed towards equilibrium |
| Time Period of Mass-Spring System | $$T = 2\pi \sqrt{\frac{m}{k}}$$ | m = mass, k = spring constant s | Massless spring, ideal Hookean regime |
| Time Period of Simple Pendulum | $$T = 2\pi \sqrt{\frac{l}{g}}$$ | l = effective length of pendulum, g = acceleration due to gravity s | Small angular amplitude |
| Total Energy of SHM | $$E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2$$ | E = total mechanical energy J | Conservative system, no damping |
| Physical Pendulum Time Period | $$T = 2\pi \sqrt{\frac{I}{mgd}}$$ | I = moment of inertia about pivot, d = distance from pivot to center of mass s | Rigid body oscillating about a fixed horizontal axis |
| Torsional Pendulum Time Period | $$T = 2\pi \sqrt{\frac{I}{C}}$$ | I = moment of inertia, C = restoring couple per unit twist s | Elastic torsion within elastic limit |
Draw: Three synchronized sinusoidal curves plotted against time t, showing phase relationships (velocity leads displacement by π/2 radians, acceleration leads velocity by π/2 radians and is in antiphase with displacement).
Interpret: Highlights the kinematic interrelationships and phase differences between displacement, velocity, and acceleration.
Exam use: Conceptual MCQs on phase relationships and reading maximum values.
Draw: Parabolic curves for potential energy symmetric about x = 0, inverted parabolic curve for kinetic energy, and a horizontal straight line representing constant total energy E at $x = \pm A$.
Interpret: Demonstrates continuous interconversion between kinetic and potential energy while total mechanical energy remains conserved.
Exam use: Numerical problems involving energy values at specific displacements (e.g., when K = U).
Draw: A straight-line graph plotting T² on the y-axis against length l on the x-axis, passing through the origin.
Interpret: The slope of the straight line equals 4π²/g, allowing experimental determination of g.
Exam use: Practical viva and analytical graph-based numerical questions.
Answer: $v_max = 0.628 m s^-1$, $a_max = 7.89 m s^-2$
Answer: x = \pm \frac{A}{\sqrt{2}}
Answer: T = 2\pi \sqrt{\frac{2L}{3g}}
Common error: SHM is the only type of periodic motion.
Correct: All SHMs are periodic, but not all periodic motions are simple harmonic (e.g., planetary motion, motion of a simple clock pendulum with large amplitude).
Common error: Acceleration is zero at the extreme positions in SHM.
Correct: Acceleration is maximum at the extreme positions and zero at the mean (equilibrium) position.
Common error: The time period of a simple pendulum depends on its mass.
Correct: The time period formula $T = 2\pi \sqrt{l/g}$ is independent of the mass of the bob.
| Feature | Simple Pendulum | Physical Pendulum |
|---|---|---|
| Definition | Point mass suspended by a massless, inextensible string | Any extended rigid body oscillating about a horizontal axis |
| Time Period Formula | $T = 2\pi \sqrt{\frac{l}{g}}$ | $T = 2\pi \sqrt{\frac{I}{mgd}}$ |
| Inertia Factor | Depends only on total mass through length $l$ | Depends on moment of inertia $I$ about the axis of rotation |
Choose an option to reveal the explanation.
Explanation: The standard differential equation governing Simple Harmonic Motion is $\frac{d^2x}{dt^2} + \omega^2 x = 0$, where $\omega$ is the angular frequency.
Explanation: Acceleration in SHM is given by $a = -\omega^2 x$. Its magnitude is directly proportional to displacement $x$, making it maximum at the extreme positions ($x = \pm A$) and zero at the mean position.
Explanation: The total mechanical energy $E$ in SHM is the sum of maximum potential energy or maximum kinetic energy, which evaluates to $E = \frac{1}{2}m\omega^2 A^2$.
Explanation: The time period of an ideal simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$, which depends solely on the effective length $l$ and acceleration due to gravity $g$ (for small amplitudes).
Explanation: In the time period formula of a physical pendulum, $I$ denotes the moment of inertia of the rigid body about the fixed horizontal axis of rotation (pivot point).
Explanation: The velocity in SHM is given by $v = \pm \omega \sqrt{A^2 - x^2}$. At the extreme positions ($x = \pm A$), the velocity becomes zero.
Explanation: If $x(0) = 0$, then $A \cos(\phi) = 0 \implies \phi = \pm \frac{\pi}{2}$. Since velocity is negative, $v(0) = -A\omega \sin(\phi) < 0$, which requires $\sin(\phi) > 0$, hence $\phi = \frac{\pi}{2}$.
Explanation: Potential energy in SHM is given by $U = \frac{1}{2}m\omega^2 x^2$, which indicates a parabolic variation with displacement $x$.
Explanation: Kinetic and potential energies fluctuate at twice the frequency of displacement and velocity because terms like $x^2$ and $\sin^2(\omega t)$ complete two cycles per oscillation cycle.
Explanation: Kater's pendulum is a reversible physical pendulum designed to measure $g$ with high precision by ensuring the time period remains unchanged when the pivot and center of oscillation are interchanged.
Explanation: Perpendicular SHMs of equal frequency and amplitude with a phase difference of $\frac{\pi}{2}$ combine to form a circular trajectory.
Explanation: In a torsional pendulum, $C$ represents the restoring torsional couple per unit angle of twist.
Explanation: The time period of an ideal mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
Explanation: Simple harmonic motion is a special type of periodic motion governed by a linear restoring force, but periodic motions can be non-harmonic (e.g., planetary motion).
Explanation: Lissajous figures are trajectory patterns produced by superimposing two mutually perpendicular simple harmonic motions.
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: Kinetic energy $K = \frac{1}{2}m\omega^2(A^2 - x^2)$ and potential energy $U = \frac{1}{2}m\omega^2 x^2$. Equating $K = U$ gives $A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = \pm \frac{A}{\sqrt{2}}.$
Explanation: We know $v_{max} = A\omega$ and $a_{max} = A\omega^2$. Dividing $v_{max}$ by $a_{max}$ gives $\frac{v_{max}}{a_{max}} = \frac{1}{\omega}$. Since $\omega = \frac{2\pi}{T}$, we get $T = \frac{2\pi v_{max}}{a_{max}}.$
Explanation: For springs in series, the reciprocal of the effective spring constant is the sum of reciprocals: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$, which simplifies to $k_{eff} = \frac{k_1 k_2}{k_1 + k_2}$.
Explanation: If length increases by $300%$, the new length $l' = l + 3l = 4l$. The new time period $T' = 2\pi \sqrt{\frac{4l}{g}} = 2T$. The percentage increase is $\frac{T' - T}{T} \times 100\% = \frac{2T - T}{T} \times 100\% = 100\%$.
Explanation: Comparing with $x = A \sin(\omega t + \phi)$, amplitude $A = 5$ and angular frequency $\omega = 2\pi$. Maximum acceleration $a_{max} = A\omega^2 = 5 \times (2\pi)^2 = 5 \times 4\pi^2 = 20\pi^2$.
Explanation: For a uniform rod pivoted at one end, moment of inertia $I = \frac{1}{3}mL^2$ and distance to center of mass $d = \frac{L}{2}$. Substituting into $T = 2\pi \sqrt{\frac{I}{mgd}}$ gives $T = 2\pi \sqrt{\frac{\frac{1}{3}mL^2}{mg(L/2)}} = 2\pi \sqrt{\frac{2L}{3g}}$.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: Total energy $E = \frac{1}{2}m\omega^2 A^2 \implies E \propto A^2$. If energy is doubled ($2E$), amplitude becomes $\sqrt{2}A$.
Explanation: Kinetic energy contains terms dependent on $\sin^2(\omega t)$, which oscillates at twice the frequency of the displacement oscillation, i.e., $2f$.
Explanation: Displacement is proportional to $\cos(\omega t + \phi)$ and velocity is proportional to $-\sin(\omega t + \phi) = \cos(\omega t + \phi + \frac{\pi}{2})$, indicating a phase difference of $\frac{\pi}{2}$.
Explanation: For springs in parallel, the effective spring constant is $k_{eff} = k_1 + k_2 = 2k$. The time period is $T = 2\pi \sqrt{\frac{m}{2k}}.$
Explanation: In the damped harmonic oscillator equation, $2b$ (or $\gamma$) represents the damping coefficient per unit mass.
Explanation: Inside an orbiting satellite, effective acceleration due to gravity $g_{eff} = 0$. Since $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$, the time period becomes infinite (the pendulum does not oscillate).
Explanation: Rewriting $x_2 = A_2 \cos(\omega t - \frac{\pi}{2})$, the phase difference between the two collinear SHMs is $\frac{\pi}{2}$. The resultant amplitude is $A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\pi/2)} = \sqrt{A_1^2 + A_2^2}$.
Explanation: Given $K = 3U \implies \frac{1}{2}m\omega^2(A^2 - x^2) = 3 \left(\frac{1}{2}m\omega^2 x^2\right) \implies A^2 - x^2 = 3x^2 \implies 4x^2 = A^2 \implies x = \pm \frac{A}{2}$.
Explanation: Since $T = 2\pi \sqrt{\frac{I}{mgd}}$, to minimize the time period $T$, the moment of inertia $I$ about the pivot must be minimized.
Explanation: Angular frequency is expressed in radians per second, making its dimensional formula $[T^{-1}]$. latter.