SHM Restoring Force Equation & Phase SHM Restoring Force Equation & Phase An oscillation is a back-and-forth motion about a central position. In many familiar systems, when you pull a body slightly away from its rest position, a force appears that tries to bring it back. If this restoring force is directly proportional to displacement and always points toward equilibrium, the motion becomes Simple Harmonic Motion (SHM). SHM is special because everything varies sinusoidally: position, velocity, and acceleration follow clean sine/cosine laws. The heart of SHM is the restoring-force idea, which leads to the compact relation a = −ω²x and the differential equation d²x/dt² + ω²x = 0. Solving this gives x(t) = A cos(ωt + φ), where A is amplitude and φ is the phase constant that fixes the starting state. With these tools, you can predict when the mass is at extremes, when velocity peaks, and how energy sloshes between kinetic and potential forms. This lesson builds intuition first, then connects each idea to the precise equations you will use in NEET problems. remember Think of a smooth spring toy: stretch right, it pulls left; push left, it pulls right. The stronger you displace it, the stronger it pulls back. That linear pull-back is what powers SHM. We meet periodic motion daily: a swinging keychain, a vibrating phone, a plucked guitar string. Not all periodic motion is SHM, but many become SHM for small departures from equilibrium because the restoring force becomes nearly linear. In that small region, the motion is governed by a single frequency set by system parameters (like spring constant and mass), independent of amplitude. This is why small oscillations are so predictable and why linear models are so powerful: you get universal formulas that work across many systems. Periodic Motion Motion that repeats after a fixed time interval (the period). Back-and-forth motion about a mean (equilibrium) position. Oscillation The position where net force is zero. In SHM, it is the center of oscillation. Equilibrium Position Restoring Force A force directed toward equilibrium, typically proportional to displacement in SHM. Maximum magnitude of displacement from equilibrium. Amplitude (A) Angular Frequency (ω) Rate at which phase changes; related to frequency by ω = 2πf and period by ω = 2π/T. Phase The angle argument of the sine/cosine that locates the oscillator’s state within a cycle. Phase Constant (φ) Initial phase at t = 0; encodes starting displacement and velocity. Period (T) and Frequency (f) T is time for one cycle; f = 1/T is cycles per second (Hz). For many systems near equilibrium, Hooke-like behavior holds: the restoring force is proportional to displacement and opposite in direction. For a spring obeying Hooke’s law, F = −kx, k is the spring constant (stiffness). For a simple pendulum at small angles, the arc displacement s ≈ Lθ and the tangential restoring force is −mg sinθ ≈ −mgθ ≈ −(mg/L) s, again linear for small θ. These linear laws produce SHM with a characteristic angular frequency ω fixed by system parameters (e.g., ω = √(k/m) for a spring–mass, or ω = √(g/L) for a simple pendulum under the small-angle approximation). Linear restoring force points toward equilibrium; k > 0. Hooke’s Law (Restoring Force) The linear restoring force derived from Hooke's law allows us to combine it with Newton's second law to obtain the core SHM equation. Combine the restoring force with Newton’s second law to relate acceleration to displacement. Since F = m a and F = -k x , we obtain m a = -k x or a = -(k/m) x . Define 2 = k/m for convenience. This gives the signature SHM relation a = - 2 x : acceleration is always directed toward equilibrium and grows in magnitude with displacement. This single statement encodes the entire sinusoidal motion. Acceleration–Displacement in SHM Restoring acceleration proportional to displacement; negative indicates direction toward equilibrium. a = - 2 x Newton’s second law with linear restoring force. Divide both sides by m. Define angular frequency squared. Characteristic SHM relation. a = - 2 x Linear restoring force: F = -k x Point mass, no damping or driving One-dimensional motion about equilibrium Acceleration is proportional to negative displacement; slope of a–x graph is −ω². The relation a = - 2 x is equivalent to the second-order linear differential equation for displacement. Since a = d 2 x dt 2 , we write d 2 x dt 2 + 2 x = 0 . This equation says: the curvature of the x(t) graph is always proportional to -x itself. Such equations have sinusoidal solutions, which is why SHM looks like clean waves in time. Differential Equation of SHM Linear, homogeneous, constant-coefficient ODE for SHM. One-dimensional motion No damping or external driving Linear restoring force proportional to x d 2 x dt 2 + 2 x = 0 d 2 x dt 2 + 2 x = 0 Start from acceleration–displacement law. Acceleration is second derivative of displacement. Rearrange to standard ODE form. Core ODE of SHM; all kinematics and energy results flow from its sinusoidal solutions. Solving the SHM differential equation gives the time dependence of position. The characteristic equation has imaginary roots ±iω, so the general real solution is sinusoidal. Two equivalent forms are common: x(t) = A ( t + ) or x(t) = A ( t + ') . Both are identical if you shift the phase. The constants A and φ come from initial conditions, such as the starting displacement x(0) and starting velocity v(0) . Once A and φ are fixed, the entire future (and past) motion is determined. A is amplitude; φ fixes where in the cycle the oscillator starts at t = 0. Displacement in SHM Solution space of linear homogeneous ODE Real-valued motion, finite A Constant ω x(t) = A ( t + ) x(t) = A ( t + ) Start with the SHM ODE. General real solution from characteristic roots ±iω. Combine constants using a phase shift identity. Standard time law for SHM; choose sine or cosine by shifting phase. Phase tells you "where you are" in the cycle. The instantaneous phase is ( t + ) . Increasing t increases phase linearly at rate ω. If = 0 and you use cosine, the motion starts at x(0) = +A (right extreme). If = radians, it starts at x(0) = -A . If you want x(0) = 0 with the mass moving toward +x, pick x = A ( t) or use cosine with = - /2 . Choosing sine vs cosine is a convenience; the physics is entirely in A, ω, and the total phase t + . Use radians for phase. Always. When your calculator is in degrees, cos(π) will not be −1. For x = A ( t + ) , ensure ω is in rad/s and t in s so that ( t + ) is in radians. tip Write x(t) = A ( t + ) and v(t) = -A ( t + ) . At t = 0, use x(0) = A to get A . At t = 0, use v(0) = -A to get -A . Square and add: x(0) 2 + ( v(0) ) 2 = A 2 . Compute = - v(0) x(0) with quadrant decided by signs of x(0) and v(0) . Finding A and φ from initial data A, φ ω = 5 rad/s x(0) = 0.06 m v(0) = 0 m, rad easy A mass executes SHM with ω = 5 rad/s. At t = 0, x(0) = 6 cm and v(0) = 0. Find A and φ in x = A cos(ωt + φ). So φ = 0 or φ = ±π. If φ = 0, cos φ = 1 ⇒ A = 0.06 m. If φ = π, cos φ = −1 ⇒ A = −0.06 m (reject negative A by convention). Use x(0) = A and v(0) = -A . Velocity and acceleration follow directly by differentiating displacement. With x = A ( t + ) , we get v = dx dt = -A ( t + ) and a = d 2 x dt 2 = -A 2 ( t + ) = - 2 x . The velocity is 90° (π/2 rad) out of phase with displacement, and acceleration is in anti-phase with displacement. Maximum speed is v = A at the mean position ( x = 0 ). Maximum acceleration is a = 2 A at the extremes ( x = A ). v leads x by 90° in sine form; a is proportional to −x. Velocity and Acceleration in SHM Key checkpoints in one cycle: at the extremes, x = A , velocity is zero, acceleration is maximum in magnitude toward the center; potential energy is maximum, kinetic zero. At the mean position, x = 0 , velocity is maximum, acceleration is zero; kinetic energy peaks and potential is minimum. These switches repeat every half-period. Understanding these positions helps you sanity-check answers and quickly estimate phases in exam problems. Right Extreme +A −ω²A (toward 0) Maximum Mean Position ±ωA (max) Maximum Minimum Left Extreme −A +ω²A (toward 0) Maximum Position Kinetic Energy Potential Energy SHM Quantities Variable Equation ( x, v, a ) Phase Relation to Displacement Max Value Location Min Value Location Velocity leads the way by a quarter, while Acceleration and Force always run exactly opposite to the Displacement. Displacement ( x ) x = A ( t + ) Reference ( 0 ) Extreme positions ( x = A ) Mean position ( x = 0 ) Velocity ( v ) v = A 2 - x 2 Leads x by 90 ( 2 rad) Mean position ( x = 0 ) Extreme positions ( x = A ) Acceleration ( a ) a = - 2 x Leads x by 180 ( rad) Extreme positions ( x = A ) Mean position ( x = 0 ) Restoring Force ( F ) F = -k x Leads x by 180 ( rad) Extreme positions ( x = A ) Mean position ( x = 0 ) Potential Energy ( U ) U = 1 2 k x 2 N/A (Scalar frequency 2 ) Extreme positions ( x = A ) Mean position ( x = 0 ) Kinetic Energy ( K ) K = 1 2 k (A 2 - x 2) N/A (Scalar frequency 2 ) Mean position ( x = 0 ) Extreme positions ( x = A ) shm quantities Acceleration in SHM is NOT constant. Do not apply constant-acceleration kinematics (s = ut + ½at²) to an entire half-cycle. Use SHM formulas instead. neet-alert The negative sign in a = −ω²x means acceleration is negative. The sign indicates direction relative to displacement, not absolute negativity. If x is positive, a is negative (toward 0); if x is negative, a is positive. Phase is an angle. A time lag Δt corresponds to a phase difference ωΔt, and a spatial shift (for waves) corresponds to kΔx. Always convert to radians. Phase φ is the same as time lag T/4 or a length shift. In a spring–mass oscillator, = k/m . When multiple springs act together, their effective stiffness k eq changes the frequency. In parallel, springs share displacement so their forces add directly, making the system stiffer; in series, the same force stretches each spring, so total extension adds and the system softens. Replacing k with k eq in = k/m immediately updates the frequency and period. Parallel: k add; Series: reciprocals add. Use k eq in ω = √(k eq/m). k = k 1 + k 2 , k series = ( 1 k 1 + 1 k 2 ) -1 Massless, linear springs Deformations within elastic limit Quasi-static combination to define k eq k = k 1 + k 2 , k series = ( 1 k 1 + 1 k 2 ) -1 Same x for both springs. Effective stiffness increases. Same F for both springs. Divide both sides by F. Effective stiffness decreases. Once you have k eq , the angular frequency is = k eq /m and T = 2 / . This is hugely useful for compound spring systems and for quick order-of-magnitude checks: doubling stiffness increases ω by √2; doubling mass decreases ω by √2. For unknown phase problems, first pin down ω from system parameters; then use initial conditions to get A and φ. medium Two springs (k1 = 100 N/m, k2 = 150 N/m) hold a 0.50 kg block in parallel on a frictionless surface. If at t = 0 the block is released from rest at x = +0.04 m, find ω and write x(t). k1 = 100 N/m k2 = 150 N/m m = 0.50 kg x(0) = +0.040 m v(0) = 0 ω and x(t) in cosine form rad/s, m Find k eq for parallel, then ω = √(k eq/m). Use x(0) = A cos φ, v(0) = −Aω sin φ. Parallel addition. Compute ω. So φ = 0 or π. Choose φ = 0 to make cos φ = 1. Displacement x Linear a–x relation uniquely identifies SHM and the value of ω from slope magnitude. custom Equilibrium +A Right extreme (max |a|) −ω²A −A Left extreme (max |a|) +ω²A Straight line through origin with slope −ω²: a = −ω² x. control dependent derived m/ s 2 Acceleration a Real systems often lose energy to friction or drag. In light damping, the oscillation continues but its amplitude decays slowly with time. The envelope of the motion is A(t) = A 0 e - b t 2m for viscous damping force F d = -b v . The instantaneous oscillation inside still looks like a cosine with a slightly reduced angular frequency d 2 - ( b 2m ) 2 . For phase questions in light damping over a few cycles, ω and φ remain the essential descriptors; the amplitude simply shrinks. In underdamped motion, amplitude decays exponentially as A = A0 e −(b t)/(2m) . A(t) = A 0 e - b t 2 m Viscous damping: F d = -b v Underdamped case: b 2 < 4 m k No external driving Damped oscillator equation. Assume exponential solution. Complex roots yield oscillation with decay. General real solution; envelope decays exponentially. A(t) = A 0 e - b t 2 m (envelope) The exponential decay A = A0 e −(b t)/(2m) holds only for viscous damping (F ∝ v) in the underdamped regime. It does NOT apply to dry friction (Coulomb friction), where amplitude drops roughly linearly per cycle. neet-alert Phase comparisons are easy once you visualize the circle of motion. Equal increments of time advance the phase by equal angles (Δθ = ωΔt). The state repeats when the phase advances by 2π. If two oscillators have the same ω but different φ, their displacements are time-shifted copies. If their ω differ slightly, they drift out of step. In problems, convert any stated lead/lag in time to phase using = t , and remember to work in radians. Where the SHM formulas apply Small oscillations about a stable equilibrium (linear restoring force). No significant damping or driving for the pure SHM forms. One-dimensional motion (or one principal mode). Use radians for all phase angles; ω in rad/s, t in s. For pendulums: small-angle approximation (θ ≲ 10°) unless otherwise specified. At eXtreme: velocity is Zero; at Mean: Acceleration is Zero. Remember: XZ–MAZ. Extreme→v=0, Mean→a=0. neet-alert Do not mix frequency f (in Hz) and angular frequency ω (in rad/s). The correct relation is ω = 2πf. Missing 2π is a classic source of wrong answers. v = -A ( t + ), a = - 2 x x(t) is differentiable and given by cosine form No change in ω with time Start with displacement. Differentiate once. Differentiate again and substitute x. v = -A ( t + ), a = - 2 x hard A 0.40 kg block is attached to two springs in series: k1 = 300 N/m and k2 = 200 N/m. It is pulled to x = 5.0 cm and released from rest at t = 0. Find: (i) k eq, (ii) ω, (iii) x(t), and (iv) the acceleration at t = 0.40 s. k eq, ω, x(t), a(0.40 s) m = 0.40 kg k1 = 300 N/m k2 = 200 N/m x(0) = 0.050 m v(0) = 0 N/m, rad/s, m, m/s² For series: 1/k eq = 1/k1 + 1/k2. Then ω = √(k eq/m). With v(0) = 0 and x(0) = +A, choose φ = 0. Harmonic sum. Compute ω. φ = 0 and A = 0.050 m. Use a = −ω²x. Use cos(6.93) ≈ 0.80. Two significant figures: −12 m/s². To solve any SHM initial-value problem efficiently: first identify the effective stiffness and mass to get ω. Second, use A 2 = x(0) 2 + ( v(0) ) 2 to find A quickly. Third, use = - v(0) x(0) with quadrant care to find φ. Finally, write x(t) , then differentiate for v(t) and a(t) if needed. Keep units consistent and phases in radians. Use this to link a and x directly and to get ω from slopes or measured accelerations. Foundation equation whose sinusoidal solutions define SHM. Most convenient working form for initial condition problems. Convert complex spring networks to a single k to compute ω and T. Envelope decay law; underdamped, viscous friction only. Common traps and edge cases: If x(0) = 0 and v(0) > 0 , cosine form implies = - /2 (or sine form with φ = 0). If x(0) = 0 and v(0) < 0 , then = + /2 . If initial data give A = 0, re-check numbers: A cannot be negative and zero amplitude means no motion. When extracting φ from tan φ, use both signs of sine and cosine to choose the correct quadrant; do not rely on arctan alone. remember Extreme positions occur when the phase (ωt + φ) equals 0, π, 2π, … for cosine form. Mean positions occur when (ωt + φ) equals π/2, 3π/2, … Harmonic Motion Simple Harmonic Motion (SHM) Oscillation with restoring acceleration proportional to −x and solutions that are sinusoidal. Maximum |x| from equilibrium. Amplitude (A) Angular Frequency (ω) Rate of phase advance; ω = 2πf = 2π/T. Instantaneous angle θ = ωt + φ locating the state in one cycle. Phase (θ) Initial phase at t = 0. Phase Constant (φ) Restoring Force Force toward equilibrium, linear in x for SHM: F = −kx. Quick Glossary Recap Summary: SHM arises when a linear restoring force acts about equilibrium. The defining kinematic relation is a = - 2 x , equivalent to the ODE d 2 x dt 2 + 2 x = 0 . Its solution x = A ( t + ) explains amplitude, period, and phase. Determine ω from system parameters (e.g., k/m ), fix A and φ from initial conditions, and then compute v and a. Keep radians and sign conventions straight to avoid NEET traps.