Chapter 14 - Oscillations

Daily life me rectilinear, projectile, uniform circular, orbital motions padhe. Inme motion certain interval ke baad repeat hota hai periodic. Par cradle rocking, swing, boat up-down river me, piston back forth steam engine me - object moves to and fro about mean. Yeh oscillatory.

Examples: sitar guitar violin vibrating strings produce sound, drum membranes diaphragm to and fro, air molecules vibrations make sound propagation possible, solid atoms vibrate about equilibrium average energy proportional temperature, AC voltage oscillates positive negative about zero mean. To describe periodic and oscillatory we need period, frequency, displacement, amplitude, phase.

Oscillation low frequency ke liye word use hota hai like branch of tree, vibration high frequency like string of instrument. Damping ke wajah se oscillating bodies eventually rest pe aate hain friction dissipative causes se, but forced oscillations me external periodic agency se continue rakh sakte hain. Large number of coupled oscillators collective oscillations manifest as waves - water, seismic, electromagnetic. Next chapter waves.

Very often body undergoing periodic motion has equilibrium position somewhere inside path. When body at this position no net external force, if left at rest remains forever. If small displacement from position, force comes into play tries to bring body back to equilibrium point, giving rise to oscillations.

Example ball placed in bowl equilibrium at bottom, displaced little performs oscillations in bowl. Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Circular motion periodic but not oscillatory. No significant difference between oscillations and vibrations, low frequency oscillation high frequency vibration.

Period and frequency: Motion repeats itself at regular intervals, smallest interval after which motion repeated is period T SI second. Reciprocal gives number repetitions per unit time frequency ν =1/T. Unit s^-1 called hertz Hz 1 Hz=1 oscillation per second. Frequency not necessarily integer. Example human heart beats 75 times per minute frequency =75/60=1.25 Hz period T=0.8 s. For vibrations quartz crystal microseconds, orbital period Mercury 88 earth days, Halley comet 76 years, convenient units of time used.

Displacement: General sense change with time of any physical property. Rectilinear motion distance from starting point as function time is position displacement. Block attached to spring other end fixed to rigid wall, convenient measure displacement from equilibrium position. For oscillating simple pendulum angle from vertical as function time may be regarded as displacement variable. Voltage across capacitor changing with time in AC circuit also displacement variable, pressure variations in sound wave, electric magnetic fields in light wave examples. Displacement variable may take positive negative values. In experiments displacement measured for different times, represented by mathematical function of time. For periodic motion function periodic in time. Simplest periodic function f(t)=A cos ωt. If argument ωt increased by integral multiple 2π radians, value remains same. Period T=2π/ω. Similarly sin ωt periodic, linear combination A sin ωt + B cos ωt also periodic with same period, can be written as D sin(ωt+φ) where D=√(A^2+B^2) tan φ = B/A? Actually A=D cos φ? Wait D sin(ωt+φ)=D sin ωt cos φ + D cos ωt sin φ, so A=D cos φ, B=D sin φ. Fourier theorem: Any periodic function can be expressed as superposition of sine cosine functions of different periods with suitable coefficients. Examples which represent periodic and nonperiodic: sin ωt + cos ωt periodic can be written √2 sin(ωt+π/4) period 2π/ω, sin ωt + cos 2ωt + sin 4ωt periodic first term period 2π/ω, second π/ω=T/2, third 2π/4ω=T/4 smallest interval after which sum repeats is T, sum periodic with period 2π/ω. e^{-ωt} not periodic decreases monotonically tends zero, log(ωt) increases monotonically diverges, not physical displacement.

Consider particle oscillating back and forth about origin of x-axis between limits +A and -A as shown. Oscillatory motion said to be simple harmonic if displacement x from origin varies with time as x(t)=A cos(ωt+φ) where A, ω, φ constants. Thus SHM not any periodic but one in which displacement is sinusoidal function of time. Fig shows positions at discrete times 0, T/4, T/2, 3T/4, T, 5T/4. Continuous plot x vs t sinusoid.

Quantities: A amplitude magnitude maximum displacement, fixed positive. ω angular frequency, φ phase constant phase angle at t=0, (ωt+φ) phase time dependent state of motion position velocity determined by it. Two SHMs may have same A and ω but different φ as shown.

Period relation: Taking φ=0 for simplicity x(t)=A cos ωt, since motion period T x(t)=x(t+T) so A cos ωt = A cos ω(t+T) cosine periodic 2π i.e. repeats when argument changes by 2π, so ωT=2π => ω=2π/T =2π ν. ω angular frequency SI rad/s, ν frequency Hz, ω is 2π times frequency. Two SHMs may have same A and φ but different ω as seen plot curve half period twice frequency.

Example which functions represent SHM and periodic but not SHM: sin ωt - cos ωt = √2 sin(ωt-π/4) SHM period 2π/ω, sin^2 ωt = 1/2 -1/2 cos 2ωt periodic period π/ω but not SHM? Actually cos 2ωt is SHM? The function contains two terms? sin^2 is periodic harmonic with equilibrium at 1/2 instead of zero. Represents SHM? Actually cos 2ωt is SHM, sum of constant + SHM not pure SHM? Need classification.

Projection of uniform circular motion on diameter follows SHM. Experiment: Tie ball to end of string make it move in horizontal plane about fixed point with constant angular speed. Observe ball sideways or front, fixing attention in plane of motion, ball appears to execute to and fro along horizontal line with point of rotation as midpoint. Alternatively observe shadow on wall perpendicular to plane circle, what we observe is motion of ball on diameter normal to direction of viewing.

Mathematically: Suppose particle P moving uniformly on circle radius A with angular speed ω anticlockwise. Initial position vector OP at t=0 makes angle φ with positive x-axis. In time t covers further angle ωt and position vector makes angle ωt+φ with +ve x-axis. Projection of OP on x-axis x(t)=A cos(ωt+φ) defining SHM. If P moves uniformly on circle, its projection P' on diameter executes SHM. Particle P and circle sometimes referred as reference particle and reference circle. Natural unit angle radian defined ratio arc to radius, angle dimensionless not always necessary mention rad when use φ or multiples conversion radian degree not similar metre centimetre. If argument trigonometric stated without units understood radian, if degree to be used must be shown explicitly sin(15°) means 15 degree but sin(15) means 15 radians. We will often drop rad as unit, whenever angle mentioned as numerical value without units taken as radians.

Example two circular motions radius, period, initial position, sense revolution indicated figures. Obtain SHM of x-projection. Case (a) at t=0 OP makes 45°=π/4 with x-axis after time t covers angle ωt=2πt/T anticlockwise and makes angle ωt+π/4 with x-axis projection x(t)=A cos(ωt+π/4)=A cos(2πt/T+π/4) SHM amplitude A period 4 s initial phase π/4. Case (b) at t=0 OP makes 90°=π/2 with x-axis after time t covers angle ωt clockwise sense and makes angle π/2 - ωt with x-axis projection x(t)=B cos(π/2 - ωt)=B sin ωt = B cos(ωt-π/2) amplitude B period 30 s initial phase -π/2 or 3π/2.

Speed of particle in uniform circular motion is angular speed ω times radius A v=ωA direction along tangent to circle at point where particle located. From geometry velocity projection P at time t is v(t)= -ωA sin(ωt+φ) negative sign shows v opposite positive x-axis. Gives instantaneous velocity of particle executing SHM where v(t)=dx/dt. Method of reference circle can be similarly used for acceleration. Centripetal acceleration of particle P in uniform circular has magnitude v²/A = ω²A directed towards centre along PO. Instantaneous acceleration of projection particle P' is then a(t)= -ω²A cos(ωt+φ)= -ω² x(t). Gives acceleration of particle in SHM same equation can be obtained directly differentiating velocity v(t) w.r.t time a(t)=dv/dt = -ω²A cos(ωt+φ). Important property acceleration proportional to displacement, for x>0 a<0 and x<0 a>0 thus whatever value of x between -A and A acceleration always directed towards centre mean position. Putting φ=0 write x(t)=A cos ωt, v(t)=-ωA sin ωt, a(t)=-ω²A cos ωt plots shown. All quantities vary sinusoidally with time only maxima differ and different plots differ in phase x varies -A to A, v varies -ωA to ωA, a from -ω²A to ω²A. W.r.t displacement plot velocity plot has phase difference π/2 and acceleration plot phase difference π i.e. opposite.

Example body oscillates SHM equation x=5 cos(2πt+π/4) SI units angular frequency ω=2π s^-1 period 1 s at t=1.5 s displacement =5 cos(3π+π/4)=5 cos(3π+π/4)= -5*0.707=-3.535 m, speed = -ωA sin(ωt+φ)= -5*2π sin(3π+π/4)= -10π*0.707? Actually sin(3π+π/4)= -sin π/4? Let's calculate: sin(3π+π/4)=sin(13π/4)=sin(π/4+3π)= -sin π/4 = -0.707, so v= -5*2π*(-0.707)=22 m/s, acceleration a=-ω² x = -(2π)²*(-3.535)=140 m/s².

Using Newton second law and expression for acceleration a=-ω² x force acting on particle mass m in SHM F(t)=ma= -m ω² x(t) i.e. F(t)= -k x(t) where k=m ω² or ω=√(k/m). Like acceleration force always directed towards mean position hence sometimes called restoring force in SHM. To summarize discussion so far SHM can be defined in two equivalent ways either by Eq displacement x=A cos(ωt+φ) or by Eq force law F=-k x. Going from displacement to force law required differentiate two times, likewise by integrating force law two times we get back displacement. Note force in Eq linearly proportional to x(t). Particle oscillating under such force therefore called linear harmonic oscillator. In real world force may contain small additional terms proportional to x^2, x^3 etc these then called nonlinear oscillators.

Example two identical springs spring constant k attached to block mass m and to fixed supports as shown. Show when mass displaced from equilibrium position either side executes SHM find period. Answer Let mass displaced small distance x to right side equilibrium. Under this situation spring left side gets elongated length equal x and right side gets compressed same length forces acting on mass F1=-k x (spring left trying pull mass towards mean) F2=-k x (spring right trying push mass towards mean) net force F=-2k x hence force proportional displacement directed towards mean therefore motion SHM period T=2π√(m/2k).

Both kinetic and potential energies of particle in SHM vary between zero and maximum. Velocity periodic function time zero at extreme positions. Kinetic energy K=1/2 m v^2 =1/2 m ω² A² sin²(ωt+φ)=1/2 k A² sin²(ωt+φ) periodic function time being zero when displacement maximum and maximum when particle at mean position. Note since sign of v immaterial in K period of K is T/2. Potential energy U of particle executing SHM: spring force F=-k x conservative force with associated potential energy U=1/2 k x² =1/2 k A² cos²(ωt+φ) also periodic with period T/2 being zero at mean and maximum at extremes.

Total energy E=U+K=1/2 k A² cos²+1/2 k A² sin²=1/2 k A²(cos²+sin²)=1/2 k A² independent of time as expected for motion under conservative force. Both K and U always positive, K never negative since proportional square speed, potential energy positive by choice of undermined constant. Both kinetic and potential energy peak twice during each period SHM. For x=0 energy kinetic at extremes x=±A all potential. In course motion between limits kinetic increases at expense of potential or vice versa. Plots K,U,E vs time and vs displacement shown. Kinetic and potential both repeat after period T/2 total energy constant at all t or x.

Example block mass 1 kg fastened to spring spring constant 50 N/m pulled distance x=10 cm=0.1 m from equilibrium at x=0 frictionless surface from rest t=0 calculate kinetic potential total energies when 5 cm away mean position. Block executes SHM angular frequency ω=√(k/m)=√(50/1)=7.07 rad/s displacement any time t x(t)=0.1 cos(7.07 t) when particle 5 cm away 0.05=0.1 cos(7.07 t) cos=0.5 sin=0.866 velocity = -A ω sin = -0.1*7.07*0.866=-0.61 m/s K=1/2 m v²=0.19 J P.E.=1/2 k x²=1/2*50*0.05*0.05=0.0625 J total E=K+PE=0.25 J same as at max displacement 1/2 k A²=0.25 J in conformity conservation energy. At maximum displacement KE zero total energy equal PE.

There are no physical examples of absolutely pure SHM. In practice systems execute SHM approximately under certain conditions. Simplest observable example small oscillations of block mass m fixed to spring which in turn fixed to rigid wall as shown Fig. Block placed on frictionless horizontal surface. If block pulled one side and released then executes to and fro motion about mean position. Let x=0 indicate position centre block when spring equilibrium positions marked -A and +A indicate maximum displacements left right mean. Springs have special properties first discovered English physicist Robert Hooke. He had shown such system when deformed subject to restoring force magnitude proportional to deformation displacement and acts opposite direction known as Hooke's law holds good for displacements small compared to length spring. At any time t if displacement block from mean position x restoring force F acting on block F(x)= -k x constant proportionality k called spring constant value governed by elastic properties spring stiff spring has large k soft small k. Eq same as force law for SHM therefore system executes SHM from Eq we have ω=√(k/m) and period T=2π√(m/k). Stiff springs high k block small mass m attached to stiff spring will have according large oscillation frequency as expected physically.

Example 5 kg collar attached to spring spring constant 500 N/m slides without friction over horizontal rod collar displaced equilibrium position by 10.0 cm released calculate period, max speed, max acceleration. Period T=2π√(m/k)=2π√(5/500)=2π/10=0.63 s. Velocity SHM v(t)=-A ω sin(ωt+φ) max speed v_max=A ω =0.1*√(500/5)=0.1*10=1 m/s at x=0. Acceleration a(t)=-ω² x(t) max acceleration a_max=ω² A = (k/m) A =500/5*0.1=10 m/s² at extremities.

It is said Galileo measured periods swinging chandelier in church by pulse beats observed motion periodic. System kind pendulum by tying piece stone to long unstretchable thread ~100 cm long suspend from suitable support free to oscillate displace stone to one side small distance and let go stone executes to and fro motion periodic period about two seconds. Shall show this periodic motion is simple harmonic for small displacements from mean. Consider simple pendulum bob mass m tied to inextensible massless string length L other end fixed support ceiling bob oscillates in plane about vertical line through support Fig shows system free-body diagram forces acting bob.

Let θ angle made by string with vertical when bob at mean position θ=0. There are only two forces acting bob tension T along string and vertical force due gravity mg force mg can be resolved component mg cosθ along string and mg sinθ perpendicular to it. Since motion bob along circle length L centre at support bob has radial acceleration (ω² L) and also tangential acceleration latter arises since motion along arc circle not uniform. Radial acceleration provided net radial force T - mg cosθ while tangential acceleration provided mg sinθ. More convenient work with torque about support since radial force gives zero torque torque about support entirely provided tangential component force τ = -L (mg sinθ) restoring torque tends to reduce angular displacement hence negative sign. By Newton's law rotational motion τ=I α where I moment inertia system about support and α angular acceleration thus I α = -m g L sinθ Or α = -(mgL/I) sinθ. For simple pendulum I=m L² so α = -(g/L) sinθ. We can simplify if assume displacement θ small sinθ can be expressed as θ - θ³/3! + θ⁵/5!... where θ in radians. For small θ sinθ≈θ. Then α = -(g/L) θ = -ω² θ where ω=√(g/L). So motion SHM small amplitude period T=2π√(L/g).

How small should amplitude be? When you perform experiment to determine time period simple pendulum teacher tells you keep amplitude small. But how small? Should angle 1° 2° or 0.5°? Or could be 10° 20° 30°? To appreciate better measure time period for different amplitudes up to large amplitudes course for large oscillations you will have error in time period about 2% at amplitude 20°, 5% at 50°, 10% at 70° and 18% at 90°. Means error in period about 2% at 20° etc. In experiment you will never be able measure T(0) because this means no oscillations even theoretically sinθ exactly equal θ only for θ=0 there will be some inaccuracy for all other values difference increases with increasing θ therefore decide how much error tolerate. No strict rule but small usually taken <10°.

Left to itself oscillation stops eventually because mechanical energy dissipated by frictional forces such motion said to be damped damping force can be represented empirical expression F_d = -b v where b constant. Motion damped system calculated from Newton's second law F_x = -k x - b v = m d²x/dt². Solution standard methods differential equations result x = A0 e^{-(b/2m) t} cos(ω' t + φ) = A0 e^{-t/2τ} cos(ω' t + φ) where A0 maximum amplitude and τ=m/b called decay time or time constant. Frequency ω' given by ω' = √(ω0² - (b/2m)²) = √(k/m - b²/4m²) where ω0=√(k/m) frequency without damping. Dashed curves correspond to x=±A where A=A0 e^{-(b/2m) t}=A0 e^{-t/2τ} envelope decays.

If damping constant b gradually increased ω' =0 at critical value bc=2m ω0. When b ≥ bc system does not oscillate: b=bc critically damped returns to equilibrium quickly, b>bc overdamped no oscillation, b

Energy of underdamped oscillator E=1/2 m ω'² A² ≈1/2 m ω0² A0² e^{-t/τ}=E0 e^{-t/τ} where E0=1/2 m ω0² A0². Q factor (quality factor) relates to fractional energy loss per cycle infinitesimal change dE = -E0 e^{-t/τ} dt /τ = -E dt/τ. If energy loss per period ΔE small can replace dE by ΔE and dt by T (also ω'≈ω0) |ΔE|/E = T/τ =2π/(ω0 τ)=2π/Q with Q factor given by Q=ω0 τ = m ω0 / b =2π / (|ΔE|/E) per cycle. Small damping high Q sharp resonance.