PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1_LxoBKPniUQ3MCoSblWccy-G3PlrxmH5/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Periodic functions. Simple harmonic motion (S.H.M.) and its equation; phase; oscillations of a spring - restoring force and force constant; energy in S.H.M. - kinetic and potential energies; Simple pendulum - derivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion. Longitudinal and transverse waves, speed of a wave. Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound PERIODIC MOTION The motion which repeats itself after a fixed interval of time, is called periodic motion, e.g., the motion of earth around sun. Oscillatory Motion If a particle moves back and forth (or to and fro) over the same path periodically then its motion is said to be oscillatory or vibratory e.g., motion of a pendulum. Periodic Function If f(t) = f(t + T) then function is periodic and T is time period. Harmonic Motion When oscillatory motion of a particle can be expressed in terms of sine or cosine functions, it is said to be a harmonic motion. SIMPLE HARMONIC MOTION When a motion can be expressed in terms of a single sine or cosine (sinusoidal) function, the motion is said to be Simple Harmonic Motion (SHM). For SHM, force ∝-(displacement) ⇒ F ∝ – x ⇒ F = – kx [Restoring Force] CHAPTER COVERS : Periodic motion Simple harmonic motion Energy in SHM Simple pendulum Oscillation of spring Superposition of SHM ⇒ a = − k x m d 2 x + dt 2 k x = 0 or m d 2 x dt 2 + ω2 x = 0 This equation represents S.H.M. Some Important Points : x = Asinωt, x = Acosωt, x = Asinωt ± Bcosωt, x = Asin2ωt, x = Acos2(ωt + φ) all satisfy the above differential equation. Therefore they all represent S.H.M. All simple harmonic motions are oscillatory but all oscillatory motions are not simple harmonic. Two particles executing S.H.M. with time periods T1 and T2 (T1 > T2) start at the same time. The particles will be in phase after n oscillations of T2 and (n – 1) oscillations of T1 so that nT2 = (n – 1)T1. Velocity and acceleration of a particle executing S.H.M. If x = A sinωt ⇒ v = dx = Aωcos ωt dt v = Aω ⎛ωt + π ⎞ or, sin⎜ ⎟ ⎝ 2 ⎠ π i.e., velocity leads displacement by 2 . (This is always true in SHM) Dependence of velocity with position v = ω A2 − x 2 Acceleration a = dv dt = − Aω2 sin ωt ⇒ a = Aω2sin(ωt + π) π i.e., acceleration leads velocity by 2 . Acceleration and displacement are in opposite phase. Dependence of acceleration with position, is a = –ω2x Various points to remember : Variation with time Variation with position At mean position t = 0 At extreme position x = ±A Displacement x = sinAω t x = 0 x = ±A Velocity v = Aω cosω t v = ± ω A2 – x2 v = ± ωA v = 0 Displacement x = Aω2 sinω t a = –ω 2x a = 0 a = ±ω2A Energy in SHM Salient points regarding energy in SHM : Oscillating Time period Frequency Displacement T f KE T/2 2f PE T/2 2f |KE ~ PE| T/4 4f Total Energy ∞ 0 KE avg = 1 mω2A2 . 4 KEmax = 1 mω2 A2 2 at mean position. KEmin = zero at extreme position. PEavg = 1 mω2 A2 . 4 PEmax = 1 mω2 A2 2 at extreme position. Both kinetic and potential energy vary parabolically with x. Constant 1/2mω 2A2 PE = KE at x If two particles start oscillating in phase then, instant at which they come to phase again, difference of their number of vibration must be integral they come to opposite phase, difference of their number of vibrations must be odd half integral. SIMPLE PENDULUM Time period of oscillation of simple pendulum of length l for small angular amplitude (< 10º) is given by T = 2π Variation in Time Period On changing various factors, T changes as : If length ‘l’ is changed, ΔT = 1 . Δl T 2 l If gravity ‘g’ is changed, ΔT = − 1 . Δg T 2 g If length of pendulum changes due to Rise of temperature, Time loss per day = Fall of temperature, Time gain per day = αΔθ × 86400 s 2 αΔθ × 86400 s 2 If a simple pendulum is taken to height h above the earth surface, loss of time per day (h << R). = h × 86400 s R (If h = 1 km, loss of time = 13.6 s/day.) Simple Pendulum in Lift Effective g = | →g + →a | (1) a T = 2π (2) a T = 2π In case of free fall a = g T → ∞, f = 0 (3) If bob is charged and placed in electric field E. Vertical field geffective = g − qE m T = 2π Field due to fixed point charge Fixed T = 2π That is, time period does not change. Horizontal field T = 2π Seconds Pendulum T = 2s = Time period of seconds pendulum. T = 2π ⇒ l = 99 cm ≈ 1 m = Length of seconds pendulum. Simple Pendulum of Length Comparable to the Radius of Earth Time period of such a pendulum is given by, T = 2π l 2 ⎛ qE ⎞2 g + ⎜ ⎟ ⎝ m ⎠ (various cases for such pendulum) When length of the pendulum is very large. That is, l >> Re ⎡ 1 ⎤ T = 2π ⎢ l → 0⎥⎦ = 60 min. = 1.4 hr. = 84.6 minute When radius of earth is very large. That is, R >> l ⇒ T = 2π 1 → 0 R OSCILLATION OF SPRING Horizontal Oscillations The spring is pulled/pushed from x = 0 to x = x0 and released. The block executes SHM Amplitude of oscillation = x0 Time period T = 2π Smooth x = 0 x = x0 Vertical Oscillations Case - I : Supported and slowly lowered Equilibrium (mean position) In equilibrium position, extension in the spring is given by, y = mg Pulled and released 0 k Now the spring is pulled by A and released, then Amplitude of oscillation = A Time period T = 2π Case - II : Support is suddenly removed l0 k y = Mg ⇒ 0 k y0 Supported 2y0 Mean position Maximum extension = 2Mg k Amplitude of oscillation = Mg k (3) T = 2π Series Combination : (1) Smooth Smooth 1 = 1 + 1 k k1 k2 or Effective spring constant, k = k1k2 k1 + k2 , T = 2π (2) 1 = 1 + 1 k k1 k2 k k ⇒ Effective spring constant, T = 2π k = 1 2 k1 + k2 Some Important Points : (Series Combination) Force developed in both the springs will be same. x1 k2 Extensions in the two springs may be different. x = k 2 1 U1 x1 k2 Energy stored may be different. U = x = k . 2 2 1 Total extension x = x1 + x2. Parallel Combination Effective spring constant, k = k1 + k2 , T = 2π M k k1 k k2 M ⇒ Smooth Effective spring constant, k = k1 + k2 T = 2π  Smooth Smooth Smooth If we cut a spring (of spring constant k) in n equal parts then spring constant of each spring becomes nk. Various Cases of Spring Mass System Time period of oscillations in various commonly asked cases : (1) T = 2π (2) T = 2π (3) T = 2π 4kx F = − 4 (4) (5) T = 2π I = Moment of inertia R = Radius T = 2π No slipping k 1 = 1 + 1 M M1 M2 Smooth where M is called reduced mass Physical Pendulum Figure shows an extended body (called physical pendulum) pivoted about point O, which is at a distance d from its centre of mass. Time period of oscillation, T = 2π I = moment of inertia of the body about pivoted point Example cases : 1. A rod of mass m and length l suspended about its end d = l 2 , I = ml 2 3 ⇒ T = 2π