SHM Energy & Spring-Mass Systems SHM Energy & Spring-Mass Systems Simple Harmonic Motion (SHM) is a steady back-and-forth motion where the restoring force pulls the system towards equilibrium and is directly proportional to displacement. A mass on a spring is the cleanest example. The key energy idea is simple: as the mass moves, energy keeps swapping between kinetic energy (motion) and elastic potential energy (spring stretch/compression), but the total mechanical energy stays constant if there is no damping. At the ends, the mass pauses momentarily and all energy is stored in the spring; at the middle, the spring is relaxed and the mass moves fastest. That rhythm—store, release, store again—makes SHM both predictable and powerful. Once you know the spring’s stiffness and the mass, you know the angular frequency, period, and how energy sits at any point in the cycle. With combinations of springs (series or parallel), the same rules hold after you replace the set by an equivalent spring constant. In vertical springs, gravity only shifts the equilibrium, not the frequency for small oscillations. And if two masses are connected by a spring, they together oscillate as if a single "reduced mass" is attached to that spring. These are the tools you need to crack most NEET SHM numericals cleanly. remember Think of a spring-mass like a savings account switching between two wallets: one wallet is kinetic energy and the other is spring potential. Money moves between them, but the total stays fixed when there’s no fee (no damping). Amplitude ( A ) Maximum displacement from equilibrium; sets the extent of motion and the energy scale via E= 1 2 kA 2 . Rate of oscillation in rad/s; for a spring-mass, = k/m . Angular frequency ( ) Spring constant ( k ) Measure of stiffness; force per unit extension. Larger k means a stiffer spring. Sum of kinetic and potential energies; stays constant in ideal SHM: E=K+U= 1 2 kA 2 . Mechanical energy ( E ) Energy of motion: K= 1 2 mv 2 ; maximized at the mean position in SHM. Kinetic energy ( K ) Elastic potential energy ( U ) Stored in a deformed spring: U= 1 2 kx 2 ; maximized at the turning points. Phase constant ( ) Sets where in the cycle the motion starts at t=0 in x=A ( t+ ) . Positions x= A where speed is zero and all energy is potential. Turning points Effective mass for two-body oscillations with a spring: = m 1m 2 m 1+m 2 . Reduced mass ( ) We will use the Cartesian sign convention with x=0 at the equilibrium of oscillation. For a horizontal spring, this is the natural length position; for a vertical spring, equilibrium is at the gravity-shifted position x 0=mg/k (measured from natural length). Energies are computed about the oscillation equilibrium so that gravity does not affect the oscillation frequency. Unless noted, springs are ideal (massless) and motion is 1D with no damping. Kinetic energy in SHM Using v 2= 2(A 2-x 2) from energy conservation. Calculates the instantaneous energy of the oscillating mass based solely on its current velocity during simple harmonic motion. Elastic potential energy Stored energy in the deformed spring. This energy represents the work done by the spring to stretch or compress it from its equilibrium position. Total mechanical energy Constant in ideal SHM. Use this to find the total mechanical energy based on the spring constant and the maximum amplitude of oscillation. No damping or driving forces (ideal oscillator). Hooke's law is valid throughout motion. Motion is 1D about equilibrium. Total energy in SHM is constant and equals 1 2 kA 2 . Hooke's law and Newton's second law. Use a=dv/dt=(dv/dx) ,dx/dt=v ,dv/dx . Integrate with respect to x . Identify the constant as total mechanical energy. At x= A , v=0 so all energy is potential. E= 1 2 kA 2= 1 2 m 2A 2 Speed–position relation Zero at |x|=A , maximum v = A at x=0 . Energy shuttles between U= 1 2 kx 2 and K= 1 2 m 2(A 2-x 2) . At the center ( x=0 ), speed is v = A and all energy is kinetic. At the ends ( x= A ), v=0 and all energy is potential. This symmetry also means the oscillator spends slightly more time near the turning points where it moves slowly. Over one full cycle in ideal SHM. Time-averaged energies Energy conservation applies if and only if non-conservative forces are absent. With light damping, E slowly decays; the instantaneous swap K U still happens within the decaying envelope. tip Acceleration is proportional to and opposite the displacement: a=- 2 x . Sets the signature linear a – x relation for SHM. a=- 2 x Hooke's law and Newton's second law. Relate acceleration to displacement. Define angular frequency. Standard SHM signature. Acceleration–displacement relation: a=- 2 x Linear restoring force F=-kx (Hooke). Mass m is constant; motion is 1D. No damping or driving. Equation of motion for ideal SHM: d 2x dt 2 + 2 x=0 with solutions in sine/cosine. Linear restoring force F=-kx . No damping or driving. 1D motion; m constant. Differential equation of SHM: d 2x dt 2 + 2 x=0 Newton's second law with Hooke's law. Divide by m . Identify angular frequency. Canonical SHM differential equation. x + 2 x=0 General solution of SHM: x=A ( t+ ) (or sine form), where A and depend on initial conditions. Solve the linear homogeneous ODE with constant coefficients. Combine into a single cosine with phase. Compact form capturing amplitude and phase. x=A ( t+ ) SHM equation x + 2 x=0 holds. Real motion; A 0 ; real. General displacement in SHM: x=A ( t+ ) For underdamped motion, the amplitude envelope decays as A=A 0 e - bt 2m while oscillating near k/m for light damping. A=A 0 e - bt 2m Equation of motion with damping. Trial solution and characteristic equation. Complex roots in underdamping; d= k m - b 2 4m 2 . Real solution; amplitude envelope decays exponentially. Envelope of oscillation amplitude. Amplitude decay in viscous damping: A=A 0 e - bt 2m Viscous damping force F d=-b ,v (linear in speed). Underdamped regime b 2<4mk so oscillations persist. 1D motion; k and b constants. neet-alert Amplitude halves with time constant A= 2m b but energy halves with E= m b . Do not mix amplitude decay and energy decay; energy decays twice as fast in the exponent. Spring–Mass Systems: Horizontal and Vertical For a mass m attached to an ideal spring of constant k on a smooth horizontal surface, the motion is SHM about x=0 (natural length). The angular frequency and period are determined only by k and m , not by amplitude: doubling A doubles energy but leaves unchanged. Frequency and period of a spring–mass Independent of amplitude in ideal SHM. tip The formulas above assume: small deformations within the elastic limit, massless spring, no friction or air resistance, and motion confined to one dimension. In a vertical spring, gravity sets a new equilibrium at x 0=mg/k relative to the spring’s natural length. If you measure displacements y from this shifted equilibrium ( y=x-x 0 ), the equation of motion is identical to the horizontal case: y + 2 y=0 with the same = k/m . Thus gravity changes the equilibrium position but not the small-oscillation frequency. Static shift in vertical spring Gravity shifts equilibrium by x 0 . Same = k/m as horizontal case. Vertical spring motion about new equilibrium Gravity changes the time period of a vertical spring. For small oscillations, gravity only shifts equilibrium by x 0=mg/k . The period remains T=2 m/k . Often, multiple springs are combined to tune stiffness. Replace them by an equivalent spring constant k eq before applying SHM formulas. In parallel the elongation is the same and forces add; in series the force is the same and elongations add. Equivalent spring constants Generalizes to many springs by summing as appropriate. To find the effective stiffness of multiple springs, add the constants for parallel arrangements or use the inverse sum for series arrangements. Total extension equals sum of extensions. Hooke's law for each spring. Add extensions. Define equivalent stiffness. k series = ( 1 k 1 + 1 k 2 ) -1 Ideal, massless, linear springs. Same tension F through series elements. Series combination: 1 k eq = 1 k 1 + 1 k 2 k eq =k 1+k 2 Each spring exerts a restoring force proportional to x towards equilibrium. For small symmetric shifts, both contributions add. Equation of motion for the mass. Identify the equivalent stiffness. Two opposite springs on a mass: k eq =k 1+k 2 Mass attached to two springs fixed at opposite walls. Small displacements; both springs stay in the linear regime. SPRINGS rule: Series → Sum of Reciprocals; Parallel → Plain Sum. Quick check: looks like resistors but swapped (Series springs behave like Parallel resistors). Now put the ideas to work. We will calculate how energy splits at a given position, how combinations of springs change period, and how two-body oscillations reduce to a single effective mass. After each formula, make sure the units and limiting cases make sense. m/s, J, J easy Speed v , kinetic energy K , and potential energy U at x=0.060 , m . m = 0.50 kg k = 200 N/m A = 0.10 m x = +0.060 m Use = k/m , v= A 2-x 2 , K= 1 2 mv 2 , U= 1 2 kx 2 . Angular frequency from k and m . Compute speed from position. Speed at x=0.060 , m . Kinetic energy. Elastic potential energy. Check: K+U=0.64+0.36=1.00 , J =E . Energy split at a given x; standard SHM numerical. A 0.50 kg block on a horizontal spring ( k=200 , N/m ) oscillates with amplitude A=0.10 , m . Find (i) the speed and (ii) K and U when x=+0.060 , m . When springs are combined, always replace them by a single k eq first, then proceed exactly as for a single spring. Energy and period formulas use k eq . Equivalent spring constant. Angular frequency (rounded to 2 s.f.). Time period. Total mechanical energy. Two ideal springs of k 1=300 , N/m and k 2=200 , N/m are in series with a 0.50 , kg mass on a smooth surface. If A=0.050 , m , find the period and total energy. Series springs with equivalent k, period, and energy. medium Equivalent stiffness k eq , period T , and total energy E . k 1 = 300 N/m k 2 = 200 N/m m = 0.50 kg A = 0.050 m For series: k eq = ( 1 k 1 + 1 k 2 ) -1 . Then = k eq /m , T=2 / , and E= 1 2 k eq A 2 . N/m, s, J Two-body oscillations with a mass–spring–mass system reduce to a single-degree problem in the center-of-mass frame: the effective inertia is the reduced mass = m 1m 2 m 1+m 2 . The spring stores energy 1 2 k x rel 2 in the relative coordinate x rel =x 1-x 2 . Effective inertia for relative motion. Reduced mass Used in classical mechanics to simplify the analysis of two-body systems (e.g., orbiting bodies) by treating the system as a single particle Identical in form to single-mass case with m . Frequency of two-mass spring system Determines the oscillation period for the effective single-degree-of-freedom system established by the reduced mass. hard , E , v 1, , and v 2, . m 1 = 0.20 kg m 2 = 0.30 kg k = 100 N/m A rel = 0.020 m Use = m 1m 2 m 1+m 2 , = k/ , E= 1 2 kA rel 2 , and CM relations v 1= m 2 m 1+m 2 x rel , v 2= m 1 m 1+m 2 x rel with x rel, = A rel . Reduced mass. Angular frequency (2 s.f.). Total energy stored in the spring. Maximum relative speed. Max speed of m 1 (towards CM). Max speed of m 2 (opposite direction). Two-mass spring oscillator via reduced mass. Masses m 1=0.20 , kg and m 2=0.30 , kg are connected by a spring of k=100 , N/m on a frictionless line. (i) Find the angular frequency of relative oscillations. (ii) If the relative amplitude is A rel =0.020 , m , find total energy and maximum speeds of each mass. rad/s, J, m/s, m/s custom K max, U = 0 at x = 0 U = E, K = 0 at x = +A +A U = E, K = 0 at x = -A -A Energy sharing in SHM as the mass moves from one turning point to the other. Potential energy is a parabola opening upward; kinetic energy is an inverted parabola. Their sum is a horizontal line at E. U(x) = 1/2 k x 2 dependent K(x) = E - U(x) derived Energy Displacement x Kinetic energy is maximum at the extremes because the spring is most stretched. At the extremes, speed is zero, so K=0 . Kinetic energy is maximum at x=0 where the spring is momentarily unstretched and v= A . At x= A : v=0 , K=0 , U=E . At x=0 : v=v = A , K=E , U=0 . If A 0 : motion collapses to equilibrium, E 0 . If k 0 (very soft spring): 0 , very slow oscillations. With light damping ( b 2 mk ): d , amplitude envelope decays as e -bt/2m , energy as e -bt/m . Boundaries and limiting cases Quick comparison for vertical vs horizontal springs (small oscillations). Equilibrium position Natural length (define x=0 there) Shifted by x0 = mg/k below natural length Angular frequency ω = √(k/m) ω = √(k/m) (same) Energy reference Elastic U = 1/2 k x 2 Use displacement y from shifted equilibrium; U = 1/2 k y 2 Effect of gravity None Shifts equilibrium only; does not change ω Feature Horizontal spring Vertical spring neet-alert For vertical springs, compute energies using displacement from the shifted equilibrium. If you mistakenly use the natural length as zero, you will add gravity’s mgx term incorrectly and get wrong U and E splits. tip Series vs parallel trap: Two opposite springs attached to a single mass act like parallel (k adds), not series. Series occurs when springs are end-to-end sharing the same tension. Time-averages over one complete cycle help in power and heating calculations (if any). Because x(t) and v(t) are sinusoidal, the average values of 2 and 2 over a period are both 1/2 , leading to equal average kinetic and potential energies. Sketch of time-average proof Write x=A ( t+ ) so v=-A ( t+ ) . Then U= 1 2 kA 2 2( t+ ) and K= 1 2 m 2A 2 2( t+ ) . Use m 2=k to make the amplitudes equal. Average over one period: 2 = 2 =1/2 . Hence U = K = 1 2 1 2 kA 2=E/2 . Velocity at any position Sign + for motion towards +x , - for towards -x . Use this relationship to verify that the speed is zero at the maximum displacement and maximum at the center. Always check units and extreme cases: if x A , v 0 ; if x 0 , v A . If a result violates these, there is likely an algebra or sign mistake. Maximum displacement from equilibrium. Amplitude Rate of oscillation in rad/s; = k/m for a spring–mass. Angular frequency Constant sum E=K+U= 1 2 kA 2 in ideal SHM. Mechanical energy Total energy = m 1m 2 m 1+m 2 for two-body oscillations. Reduced mass k eq k eq Equivalent spring constant k parallel =k 1+k 2 , k series = ( 1 k 1 + 1 k 2 ) -1 . Key terms recap