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Oscillations Formulas For JEE 2027
Oscillations in JEE Physics covers periodic motion, with a focus on Simple Harmonic Motion (SHM). In SHM, the restoring force is proportional to displacement and acts towards the equilibrium position. Key concepts include displacement, velocity, acceleration, time period, frequency and energy, along with applications involving spring–mass systems and simple pendulums. For JEE 2027 preparation, use a JEE Mains Formula sheet to revise these equations and understand the conditions under which they apply.
This oscillations formula sheet brings together important SHM equations, energy relations and concepts such as damped oscillations, forced oscillations, resonance and the superposition of SHMs. After revising each concept, solve JEE Mains previous papers to practise identifying the right formula and applying it to exam questions. Reviewing mistakes and checking signs, units and assumptions can help improve your accuracy and problem-solving speed.
Periodic and Oscillatory Motion: Definitions
Many motions in nature repeat themselves over and over — the swinging of a pendulum, the vibration of a guitar string, the bobbing of a boat on waves. Any motion that repeats at regular intervals of time is called periodic motion. When this periodic motion is a back-and-forth movement about a central position, it is called oscillatory motion.
Periodic Motion: Motion that repeats itself after a fixed interval of time called the time period ($$T$$)
Oscillatory Motion: A special type of periodic motion where an object moves back and forth about a fixed point called the equilibrium position (or mean position)
Time Period ($$T$$): The time taken for one complete cycle of oscillation. Measured in seconds (s).
Frequency ($$f$$): The number of complete oscillations per second. $$f = 1/T$$. Measured in hertz (Hz).
Angular Frequency ($$\omega$$): The rate of oscillation in radians per second. $$\omega = 2\pi f = 2\pi/T$$.
Simple Harmonic Motion (SHM)
Simple Harmonic Motion is the simplest and most important type of oscillatory motion. Imagine pulling a spring and releasing it — the mass oscillates back and forth. This back-and-forth motion, where the restoring force is proportional to the displacement from the mean position, is SHM.
Simple Harmonic Motion (SHM): An oscillatory motion in which the restoring force (or acceleration) is directly proportional to the displacement from the mean position and is always directed towards it
Condition for SHM
The restoring force must satisfy:
$$$F = -kx$$$
or equivalently, the acceleration:
$$$a = -\omega^2 x$$$
where $$x$$ = displacement from mean position, $$k$$ = force constant, $$\omega^2 = k/m$$.
The negative sign means the force/acceleration always points towards the mean position (opposes displacement).
Note: The key test for SHM: if you can write $$a = -\omega^2 x$$ (acceleration proportional to displacement with a negative sign), the motion is SHM with angular frequency $$\omega$$.
Displacement, Velocity, and Acceleration in SHM
The position of an object in SHM varies sinusoidally with time. From this, we can derive expressions for velocity and acceleration at any instant.
Equations of SHM
Displacement:
$$$x(t) = A\sin(\omega t + \phi)$$$
Velocity:
$$$v(t) = A\omega\cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}$$$
Acceleration:
$$$a(t) = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x$$$
where:
- $$A$$ = amplitude (maximum displacement from mean position)
- $$\omega$$ = angular frequency
- $$\phi$$ = initial phase (depends on starting conditions)
- $$(\omega t + \phi)$$ = phase at time $$t$$
Key observations about these equations:
- At mean position ($$x = 0$$): velocity is maximum ($$v = \pm A\omega$$), acceleration is zero
- At extreme position ($$x = \pm A$$): velocity is zero, acceleration is maximum ($$a = \mp A\omega^2$$)
- Velocity leads displacement by $$\pi/2$$ (a quarter cycle ahead)
- Acceleration leads velocity by $$\pi/2$$ (or is $$\pi$$ out of phase with displacement)
Worked Example
A particle in SHM has amplitude 5 cm and time period 0.2 s. Find: (a) the maximum velocity, (b) the maximum acceleration, (c) the velocity when displacement is 3 cm.
$$A = 5$$ cm $$= 0.05$$ m, $$T = 0.2$$ s
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi$$ rad/s
(a) $$v_{\max} = A\omega = 0.05 \times 10\pi = 0.5\pi \approx 1.57$$ m/s
(b) $$a_{\max} = A\omega^2 = 0.05 \times (10\pi)^2 = 0.05 \times 100\pi^2 = 5\pi^2 \approx 49.3$$ m/s$$^2$$
(c) $$v = \omega\sqrt{A^2 - x^2} = 10\pi\sqrt{0.05^2 - 0.03^2} = 10\pi\sqrt{0.0016} = 10\pi \times 0.04 = 0.4\pi \approx 1.26$$ m/s
Time Period and Frequency
Time Period of SHM
$$$T = 2\pi\sqrt{\frac{m}{k}} = \frac{2\pi}{\omega}$$$
where $$m$$ = mass of the oscillating object, $$k$$ = force constant (spring constant).
Equivalently:
$$$T = 2\pi\sqrt{\frac{\text{displacement}}{\text{acceleration}}} = 2\pi\sqrt{\frac{x}{|a|}}$$$
(This is very useful for finding $$T$$ in complex systems.)
Tip: For any SHM problem, the strategy is: (1) find the net restoring force, (2) write it in the form $$F = -kx$$, (3) identify $$\omega^2 = k/m$$, and (4) compute $$T = 2\pi/\omega$$.
Energy in SHM: Kinetic and Potential
A particle in SHM continuously converts energy between kinetic energy (when moving fast near the mean position) and potential energy (when stretched/compressed at the extremes). The total energy remains constant.
Energy in SHM
Kinetic Energy:
$$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}kA^2\cos^2(\omega t + \phi)$$$
Potential Energy:
$$$PE = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(\omega t + \phi)$$$
Total Energy:
$$$E = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2 = \text{constant}$$$
Key energy facts:
- Total energy is proportional to $$A^2$$ (double the amplitude $$\rightarrow$$ quadruple the energy)
- At $$x = 0$$ (mean position): all energy is kinetic, $$KE = E$$, $$PE = 0$$
- At $$x = \pm A$$ (extremes): all energy is potential, $$PE = E$$, $$KE = 0$$
- At $$x = \pm A/\sqrt{2}$$: energy is equally split, $$KE = PE = E/2$$
- Both KE and PE oscillate with frequency $$2\omega$$ (twice the SHM frequency)
Worked Example
A 0.5 kg mass on a spring ($$k = 200$$ N/m) oscillates with amplitude 10 cm. Find: (a) total energy, (b) maximum speed, (c) speed when $$x = 6$$ cm.
(a) $$E = \frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1$$ J
(b) At mean position, all energy is kinetic:
$$\frac{1}{2}mv_{\max}^2 = E \Rightarrow v_{\max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 1}{0.5}} = 2$$ m/s
(c) $$v = \omega\sqrt{A^2 - x^2}$$, $$\omega = \sqrt{k/m} = \sqrt{200/0.5} = 20$$ rad/s
$$v = 20\sqrt{0.1^2 - 0.06^2} = 20\sqrt{0.0064} = 20 \times 0.08 = 1.6$$ m/s
Tip: JEE shortcut: At what displacement is $$KE = PE$$? Set $$\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$$, giving $$x = A/\sqrt{2}$$. At what displacement is $$KE = nPE$$? Then $$x = A/\sqrt{n+1}$$.
Spring-Mass System Formulas
A mass attached to a spring is the most common physical example of SHM. When the spring is stretched or compressed from its natural length, it exerts a restoring force described by Hooke's law.
Hooke's Law: The restoring force of a spring is proportional to its extension or compression: $$F = -kx$$, where $$k$$ is the spring constant (in N/m)
Spring-Mass System
Horizontal spring:
$$$T = 2\pi\sqrt{\frac{m}{k}}, \quad \omega = \sqrt{\frac{k}{m}}$$$
Vertical spring: Same formula for $$T$$ and $$\omega$$. The mean position shifts by $$x_0 = mg/k$$ (the static extension), but the time period is unaffected by gravity.
Combinations of Springs
When multiple springs are connected, we need the effective spring constant to find the time period.
Springs in Series and Parallel
Parallel (side by side, both stretch the same amount):
$$$k_{\text{eff}} = k_1 + k_2 + k_3 + \cdots$$$
Series (end to end, same force through both):
$$$\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \cdots$$$
For two springs in series: $$k_{\text{eff}} = \dfrac{k_1 k_2}{k_1 + k_2}$$
Worked Example
A 2 kg block is connected to two springs ($$k_1 = 300$$ N/m, $$k_2 = 600$$ N/m) in parallel. Find the time period of oscillation.
$$k_{\text{eff}} = k_1 + k_2 = 300 + 600 = 900$$ N/m
$$T = 2\pi\sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi\sqrt{\frac{2}{900}} = 2\pi\sqrt{0.00222} = 2\pi \times 0.0471 = 0.296$$ s
Worked Example
The same two springs are now connected in series with the same block. Find the new time period.
$$k_{\text{eff}} = \frac{k_1 k_2}{k_1 + k_2} = \frac{300 \times 600}{300 + 600} = \frac{180000}{900} = 200$$ N/m
$$T = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1 = 0.628$$ s
The series combination gives a longer time period (softer effective spring).
Simple Pendulum Formula
A simple pendulum is a small heavy bob suspended by a light, inextensible string from a fixed point. When displaced slightly and released, it oscillates back and forth. For small angles (less than about $$15^\circ$$), the motion is approximately SHM.
Simple Pendulum
$$$T = 2\pi\sqrt{\frac{l}{g}}$$$
where $$l$$ = length of the pendulum (from pivot to centre of bob), $$g$$ = acceleration due to gravity.
- Valid only for small angular amplitudes ($$\theta_0 \ll 1$$ radian)
- $$T$$ is independent of the mass of the bob
- $$T$$ is independent of the amplitude (for small oscillations)
- $$\omega = \sqrt{g/l}$$
Worked Example
Find the length of a simple pendulum with a time period of 2 s (a "seconds pendulum") on Earth ($$g = 9.8$$ m/s$$^2$$).
$$T = 2\pi\sqrt{\frac{l}{g}} \Rightarrow l = \frac{gT^2}{4\pi^2}$$
$$l = \frac{9.8 \times 4}{4\pi^2} = \frac{39.2}{39.48} = 0.993$$ m $$\approx 1$$ m
Pendulum in Special Situations
Effective $$g$$ in Different Situations
Replace $$g$$ with $$g_{\text{eff}}$$ in the pendulum formula:
- In a lift accelerating upward with acceleration $$a$$: $$g_{\text{eff}} = g + a$$ (shorter $$T$$)
- In a lift accelerating downward with acceleration $$a$$: $$g_{\text{eff}} = g - a$$ (longer $$T$$)
- In free fall ($$a = g$$): $$g_{\text{eff}} = 0$$ (pendulum doesn't oscillate)
- In a horizontal acceleration $$a$$ (e.g., a car): $$g_{\text{eff}} = \sqrt{g^2 + a^2}$$
Tip: If a pendulum question says "at the Moon" or "on a mountain," just change $$g$$ accordingly. Time period $$T \propto 1/\sqrt{g}$$, so if $$g$$ decreases, $$T$$ increases (pendulum swings slower).
Compound Pendulum Formula
A compound pendulum is any rigid body that can oscillate about a horizontal axis that does not pass through its centre of mass. Unlike a simple pendulum, the mass is distributed along the body.
Compound Pendulum: A rigid body oscillating about a fixed horizontal axis under gravity. Examples: a swinging door, a metronome, a uniform rod pivoted at one end.
Compound Pendulum
$$$T = 2\pi\sqrt{\frac{I}{mgd}}$$$
where:
- $$I$$ = moment of inertia about the pivot axis
- $$m$$ = mass of the body
- $$g$$ = acceleration due to gravity
- $$d$$ = distance from the pivot to the centre of mass
Worked Example
A uniform rod of length $$L$$ and mass $$m$$ is pivoted at one end. Find the time period of small oscillations.
Moment of inertia about the end: $$I = \frac{mL^2}{3}$$
Distance from pivot to centre of mass: $$d = L/2$$
$$T = 2\pi\sqrt{\frac{I}{mgd}} = 2\pi\sqrt{\frac{mL^2/3}{mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$$
Damped Oscillations Formulas
In the real world, oscillations don't go on forever. Friction, air resistance, and other dissipative forces gradually reduce the amplitude over time. These are called damped oscillations.
Damping: The gradual decrease in amplitude of oscillation due to dissipative forces (like friction or viscosity)
Damped SHM
If the damping force is proportional to velocity ($$F_{\text{damp}} = -bv$$), the displacement is:
$$$x(t) = A_0 \, e^{-bt/2m} \sin(\omega' t + \phi)$$$
where:
- $$A_0 \, e^{-bt/2m}$$ = amplitude that decreases exponentially with time
- $$b$$ = damping constant
- $$\omega' = \sqrt{\omega_0^2 - \left(\frac{b}{2m}\right)^2}$$ = damped angular frequency
- $$\omega_0 = \sqrt{k/m}$$ = natural (undamped) angular frequency
The energy of the oscillator decreases as:
$$$E(t) = \frac{1}{2}kA_0^2 \, e^{-bt/m}$$$
Three regimes of damping:
- Underdamped ($$b < 2m\omega_0$$): the system oscillates with decreasing amplitude — the most common case
- Critically damped ($$b = 2m\omega_0$$): the system returns to equilibrium in the shortest time without oscillating
- Overdamped ($$b > 2m\omega_0$$): the system returns to equilibrium slowly without oscillating
Forced Oscillations and Resonance Formulas
If you push a swing at regular intervals, you can make it swing higher and higher. This is an example of forced oscillation — an external periodic force drives the system. The amplitude of the response depends on how close the driving frequency is to the system's natural frequency.
Forced Oscillation: An oscillation maintained by an external periodic driving force $$F = F_0\sin(\omega_d t)$$
Resonance: The phenomenon where the amplitude of a forced oscillation becomes maximum when the driving frequency equals (or is close to) the natural frequency of the system
Forced Oscillations
When a periodic force $$F_0\sin(\omega_d t)$$ is applied to a damped oscillator:
Steady-state amplitude:
$$$A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + \left(\frac{b\omega_d}{m}\right)^2}}$$$
- At resonance ($$\omega_d \approx \omega_0$$): amplitude is maximum
- With low damping, the resonance peak is sharp and tall
- With high damping, the resonance peak is broad and short
- The oscillation frequency equals the driving frequency $$\omega_d$$ (not $$\omega_0$$)
Note: At resonance, maximum energy transfer occurs from the driving force to the oscillator. This explains why soldiers break step on a bridge — marching in step could drive the bridge at its natural frequency, causing dangerously large oscillations.
Superposition of Two SHMs
When two SHMs act on the same particle simultaneously, the resultant displacement is the sum of the individual displacements. The result depends on the amplitudes, frequencies, and the phase difference between them.
Same Frequency, Same Direction
Superposition of Two SHMs (Same Frequency)
If $$x_1 = A_1\sin(\omega t)$$ and $$x_2 = A_2\sin(\omega t + \phi)$$, the resultant is:
$$$x = A\sin(\omega t + \delta)$$$
where the resultant amplitude:
$$$A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos\phi}$$$
- $$\phi = 0$$ (in phase): $$A = A_1 + A_2$$ (constructive, maximum)
- $$\phi = \pi$$ (opposite phase): $$A = |A_1 - A_2|$$ (destructive, minimum)
- $$\phi = \pi/2$$: $$A = \sqrt{A_1^2 + A_2^2}$$
Worked Example
Two SHMs of amplitudes 3 cm and 4 cm with a phase difference of $$\pi/2$$ are superposed. Find the resultant amplitude.
$$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\pi/2)}$$
$$= \sqrt{9 + 16 + 0} = \sqrt{25} = 5$$ cm
Slightly Different Frequencies (Beats in SHM)
When two SHMs of slightly different frequencies are superposed, the resultant amplitude varies slowly — it waxes and wanes periodically. This is the SHM analog of "beats" in sound.
Superposition of Slightly Different Frequencies
If $$x_1 = A\sin(\omega_1 t)$$ and $$x_2 = A\sin(\omega_2 t)$$ with $$\omega_1 \approx \omega_2$$:
$$$x = 2A\cos\left(\frac{\omega_1 - \omega_2}{2}t\right)\sin\left(\frac{\omega_1 + \omega_2}{2}t\right)$$$
The amplitude varies with frequency $$|\omega_1 - \omega_2|/2$$, and the beat frequency is $$|f_1 - f_2|$$.
Key Formulas at a Glance
Quick Reference
- SHM condition: $$a = -\omega^2 x$$
- Displacement: $$x = A\sin(\omega t + \phi)$$
- $$v = \omega\sqrt{A^2 - x^2}$$, $$v_{\max} = A\omega$$, $$a_{\max} = A\omega^2$$
- $$T = 2\pi/\omega = 2\pi\sqrt{m/k}$$
- Total energy: $$E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$$
- Simple pendulum: $$T = 2\pi\sqrt{l/g}$$
- Compound pendulum: $$T = 2\pi\sqrt{I/(mgd)}$$
- Springs in parallel: $$k_{\text{eff}} = k_1 + k_2$$
- Springs in series: $$1/k_{\text{eff}} = 1/k_1 + 1/k_2$$
- Superposition amplitude: $$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$$
Oscillations Formulas For JEE 2027: Conclusion
Oscillations is one of the JEE Mains Physics Important chapters to revise thoroughly, particularly for concepts such as SHM, spring–mass systems, pendulums and energy relations. Understand how displacement, velocity and acceleration vary during SHM, and check the assumptions behind each formula before applying it to a problem.
Use a JEE Physics formulas sheet to revisit key equations during practice and last-minute revision. Combine formula revision with numerical and conceptual questions, review your mistakes, and pay attention to units and signs. This approach can help build accuracy, improve problem-solving speed and strengthen your preparation for JEE Main and Advanced 2027
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