Tiêu đề 6. Simple Harmonic Motion.pdf ✅

112 Quick Revision NCERT-PHYSICS Figure: 6.6 m T 2 g = π l ( ) 2 0 v g = − 2 cos cos l θ θ String tension tension T mg = − (3cos 2cos ) θ θ Second pendulum has time period = 2 second length nearly 1 m (on earth). If length of pendulum is sufficient, ( ) 2 1 / R T g R l π     =     +   If l → ∞, 2 84.6 R T g = = π min Pendulum in an Accelerated Vehicle: The time period of simple pendulum whose point of suspension moving horizontally with acceleration a In this case effective acceleration 2 2 eff . g g a = + 2 2 1/ 2 2 ( ) l T g a = π + and 1 tan a g θ − = If simple pendulum suspended in a car that is moving with constant speed v around a circle of radius r. 2 2 2 2 l T v g r = π   +     Elastic (spring) force F = – kx applied to mass m 0 2 2 m T k π π ω = = 1 2 2 E k x p = , 1 2 2 E k A = When a spring of length l is cut in two pieces of length l1 and l2 such that = nll 21 . If the constant of a spring is k then spring constant of first part 1 k n( 1) k n + = Spring constant of second part 2 k n k = + ( 1) and ratio of spring constant 1 2 k 1 k n = If second pendulum is taken at height ; h R << e time period increase and clock loses time given by 86400 e e h h n R R × = × per day If second pendulum is taken at depth ; h R << e the clock loses time given by 864000 2 2 e e h h n R R × = × per day If length of second pendulum is increased by ∆l, the clock loses time given by 86400 2 2 l l n l l ∆ ∆ × = × per day If length of second pendulum is decreased, the clock gains time given by 86400 2 2 l l n l l ∆ ∆ × = × per day Damping reduces the frequency of oscillators. Conical Pendulum: The force acting on the bob are tension and weight of the bob. From the figure 2 sin mv T r θ = and T mg cosθ = Figure: 6.7 Tension in the string: 2 2 1 v T mg rg   = +     2 2 cos mg mgl T θ l r = = − [As 2 2 cos h l r l l θ − = = ] Angle of string from the vertical: 2 tan v rg θ = Linear velocity of the bob: v gr = tanθ Angular velocity of the bob: tan g r ω θ = cos g g r l θ = = Time period of revolution: cos P 2 2 l h T g g θ = = π π 2 2 2 2 tan l r r g g π π θ − = = O θ P r l h S O θ S θ T cosθ T P mg T sinθ mv2 /r Figure: 6.5 a θ a θ g geff.
Simple Harmonic Motion 113 Torsional pendulum 0 2 I T D = π Restoring torque for a circular wire of radius r and length l, being twisted an angle φ 4 2 r M G D π = = φ φ l G = shear modulus of wire, D = torque constant Compound pendulum: Any rigid body suspended from a fixed support constitutes a physical pendulum. Consider the situation when the body is displaced through a small angle θ. Torque on the body about O is given by τ θ = mgl sin . . .(i) where l = distance between point of suspension and centre of mass of the body. If I be the M.I. of the body about O. Then τ α = I . . .(ii) 2 2 2 d dt θ = −ω θ we get, 2 mgl I T I mgl ω π = ⇒ = Also 2 cm I I ml = + (Parallel axis theorem) 2 2 = + mk ml (where k = radius of gyration) 2 2 2 2 2 K l mK ml l T mgl g π π + + = = eff 2 l g = π leff = Effective length of pendulum = Distance between point of suspension and centre of mass. Figure: 6.9 Longitudinal oscillations of an elastic wire: Wire/string pulled a distance ∆l and left. It executes longitudinal oscillations. Restoring force l F AY l   ∆ = −     Y = Young’s modulus; A = Area of cross-section Hence 2 2 m ml T k AY = = π π S.H.M. of a small ball rolling down in hemi-spherical bowl 2 R r T g π − = R = Radius of the bowl r = Radius of the ball Superposed oscillations: 1 1 1 x A t = sinω , 2 2 2 x A t = sinω Amplitude for 1 2 x x + ( ( ) ) 1/ 2 2 2 A A A A A t = + + − 1 2 1 2 1 2 2 cos ω ω In particular, if A A 1 2 = , 1 1 2 1 2 cos ( ) 2 A A t = − ω ω Coupled oscillations m x k x k x x 1 1 1 1 1 2 ɺɺ = − − − ( ) m x k x k x x 2 2 2 2 1 2 ɺɺ = − − − ( ) ( )2 2 2 1 1 2 2 1 2 1 1 1 2 2 2 E k x k x k x x p = + + − Solution for 1 2 k k = and m m m 1 2 = = x A t A t 1 1 1 1 2 2 2 = + + + sin sin (ω φ ω φ ) ( ) x A t A t 2 1 1 1 2 2 2 = + − + sin sin (ω φ ω φ ) ( ) 1 1 k m ω = 1 2 2k k m ω + = Free Oscillation with Damping Figure: 6.12 1.0 Overdamped The underdamped response of the oscillator Is described by the equation: x = e –γt A cos [ω1t – α] Critical damping Oscillator with resonant, Frequency 10 rad/s started, From rest. After Barger and Olsson One-half of critical damping One-tenth of critical damping 0 x 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.4 0.6 0.8 1.0 1.2 1.4 1.6 t G θ mg l CM O I I0 Figure: 6.8 θ Disc Wire l R Figure: 6.11 l Wir ∆l Figure: 6.10
114 Quick Revision NCERT-PHYSICS Differential equation 2 2 0 0 ɺɺ ɺ x x x + + = γ ω Figure: 6.13 Solution for the case of weak damping, 0 γ ω< cos( ) t e x Ae t γ ω α − = − ( 1 2 cos sin ) t e e x e A t A t γ ω ω − = − ( 1 2 ) e e t i t i t x e A e A e −γ ω ω− = + Eigen frequency, 2 2 ω ω γ e = −0 Energy of oscillation 2 k k 0 E E − γ t = Solution in the case of strong damping, 0 γ ω> ( ) ' ' 1 2 t t e e t x e A e A e −γ ω ω = + 2 2 ω γ ω e 0 ′ = − Solution in the case of critical damping, ( 1 2 ) t x e A A t −γ = + Free oscillations without damping (harmonic oscillator) Differential equation 2 ɺɺx x + = ω0 0 Solution x A t = + cos(ω φ 0 ) 1 0 2 0 x A t A t = + cos sin ω ω ( ) 0 i t x Ae ω φ+ = Forced Oscillations Differential equation 2 0 2 0 F i t z yz z e m ω ɺɺ ɺ + + = ω Particular solution z A t = − cos(ω φ ) ; ( ) i z Ae t = − ω φ Amplitude of oscillation and phase displacement ( ) ( ) ( ) 0 0 2 2 2 2 2 2 0 0 2 F m A A γ ω ω γω ω ω γ = = − + − + 2 2 0 2 tan γω φ ω ω = − Resonance amplitude, small damping, 0 0 2 0 F A mω γ = Resonance amplification (sharpness of resonance, quality), 0 2 Q ω γ = Resonance frequency 2 2 0 0 2 ω ω γ ω r = − ≈ Resonance frequency when amplitude of applied function is proportional to ω (e.g. electric current in a series resonance circuit) ω ω r = Composition of two SHM: Along the same line 1 y a wt = sin and y b wt 2 = + sin ( φ ) ( ) 1 1 2 y y y a wt e = + = + sin Where 1 2 2 a a b abc = + + 2 cosφ and 1 sin tan cos b e a b φ φ −   =     + At interagency to each other x a wt = sin and y b wt = + sin( φ ) 2 2 2 2 2 2 cos sin y x xy b a ab + − = φ φ At right angles to each other where the frequencies ratios 1: 2 x a wt and y b wt = = + sin sin 2( φ ) 2 2 2 2 2 4 sin sin 1 0 y x x y b a a b φ φ       − + + − =       Figure: 6.14 Resonance Amplitude of forced oscill Amplitude of drivi ng frequency of driving force f0 (natural frequency) Zero damping Light damping Medium damping Heavy damping