Class 11 Physics · Chapter 13
Short answer:
Class 11 Physics Chapter 13 Oscillations ke NCERT solutions yahan hain — saare 25 exercise questions ka step-by-step hal (spring-mass system, simple pendulum, damped oscillations, reference circle method), formulas ki poori list, aur exam me sabse zyada hone wali galtiyan. Ye chapter periodic motion, SHM ka force law (F = −kx), energy conservation, aur resonance cover karta hai — jo class 11 physics ncert solutions ke poore set me Chapter 14 Waves ki seedhi neev banta hai.
Chapter 13 Oscillations Class 11 Physics ka wo chapter hai jo periodic aur oscillatory motion ki poori "language" define karta hai — amplitude, time period, frequency, angular frequency, phase — jo aage Chapter 14 Waves (jahan Doppler Effect formula tak yahi language use hoti hai) aur Class 12 ke AC circuits, sound, resonance tak har jagah repeat hoti hai. Jaise Chapter 1 poore Class 10 Chemistry ki neev hai, waise hi Oscillations poore "vibrating systems" ke concept ki neev hai.
2026-27 ke rationalised syllabus (class 11 physics deleted syllabus 2026-27) me purana standalone "Physical World" chapter examinable nahi raha — syllabus seedha Units and Measurements se shuru hota hai, aur is naye 14-chapter structure me Oscillations Chapter 13 hai, System of Particles and Rotational Motion, Gravitation, aur Kinetic Theory of Gases ke baad, Waves se theek pehle. (Note: ye chapter-list registry/aggregator sources se cross-verify ki gayi hai, ncert.nic.in ka official PDF is run me directly access nahi ho paya — numerically precise cheez jaise marks weightage publish hone se pehle apne school ke latest CBSE circular se ek baar match kar lena.)
Agar tum poora class 11 physics ncert solutions series follow kar rahe ho — Motion in a Straight Line (class 11 physics chapter 2 motion in a straight line ncert solutions), motion in a plane projectile motion class 11 solutions, units and measurements class 11 important questions, laws of motion class 11 ncert exemplar, gravitation class 11 numericals with solutions, system of particles and rotational motion important formulas, thermodynamics class 11 physics notes, kinetic theory of gases class 11 derivation — to Oscillations wahi jagah hai jahan pehli baar "restoring force displacement ke directly proportional aur opposite direction me hai" (F = −kx) formally set hota hai. Yehi idea age mechanical properties of fluids class 11 numericals ke buoyancy-oscillation problems (cork ka U-tube wala sawaal isi chapter me hai) aur waves class 11 physics doppler effect formula tak carry hota hai.
Chapter 13 Summary — 5 Minute Revision
Oscillations chapter ka core hai: SHM sirf tab hoti hai jab acceleration displacement ke directly proportional aur mean position ki taraf ho (a = −ω²x). Isi ek condition se poora chapter — spring-mass system, simple pendulum, damped/forced oscillation, resonance — nikalta hai. Numericals me 3 skills baar-baar test hoti hain: (1) ω = √(k/m) ya √(g/l) nikalna, (2) x(t), v(t), a(t) equations likhna initial conditions se, (3) reference-circle method se SHM ko circular motion ke projection ke roop me samajhna.
Oscillations class 11 physics important questions me numerical zyada aate hain (spring constant, time period, max velocity/acceleration) — isliye formula table ko ratt kar nahi, derive karke yaad rakho. Agla step: Chapter 14 Waves, jahan yehi SHM equations ek travelling wave y(x,t) = A sin(kx − ωt) me convert ho jaati hain.
In-Text Questions — Solutions
NCERT Physics ke chapters me Chemistry jaisa 'in-text questions' section nahi hota — sirf chapter ke end me Exercises hote hain. Phir bhi, quick self-check ke liye: Periodic motion aur Oscillatory (vibratory) motion me kya fark hai?
Har oscillatory motion periodic hoti hai (fixed time interval T baad khud ko repeat karti hai), lekin har periodic motion oscillatory nahi hoti. Jaise Earth ka apni axis pe rotation periodic hai (24 ghante me repeat) lekin ye 'to and fro' motion nahi hai, isliye oscillatory nahi. Oscillatory motion me particle ek mean (equilibrium) position ke dono taraf baar-baar aata-jaata hai — jaise pendulum ya spring-mass system.
SHM (Simple Harmonic Motion) hone ke liye do zaroori conditions kya hain?
(i) Motion periodic honi chahiye, aur (ii) restoring force (ya acceleration) displacement x ke directly proportional honi chahiye aur opposite direction me act karni chahiye:
F = −kx ⟹ a = −ω²x
Agar force x ke square, cube, ya kisi non-linear power ke proportional hai to motion periodic to ho sakti hai, SHM nahi.
Angular frequency ω aur time period T ka relation likho.
ω = 2π/T = 2πf
Yahan f = frequency (Hz) = oscillations per second, T = time period (s) = ek complete oscillation ka time.
SHM me total mechanical energy (KE + PE) time ke saath constant kyun rehti hai?
Kyunki spring-mass ya pendulum system me sirf conservative restoring force kaam kar raha hai, koi friction/damping nahi (ideal case). Jaise particle mean position se extreme ki taraf jaata hai, KE ghatti hai lekin utni hi PE badhti hai — total E = ½kA² hamesha same rehta hai. Ye energy conservation ka direct application hai.
Forced oscillation me resonance kab hoti hai?
Jab external driving force ki frequency, system ki natural frequency ke barabar (ya bahut close) ho jaati hai, tab amplitude bahut zyada badh jaati hai — isse resonance kehte hain. Real systems me damping ki wajah se amplitude infinite nahi hoti, lekin sabse zyada zaroor hoti hai.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q25)
13.1 Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
(a) Not periodic. Swimmer har baar same exact time me nahi repeat karta — motion swimmer ke effort par depend karti hai, aur ek complete trip ke baad ruk jaati hai, kisi fixed time interval me continuously repeat nahi hoti.
(b) Periodic. Magnet apni N-S equilibrium direction ke around oscillate karta hai (Earth ke magnetic field ki restoring torque ki wajah se) — fixed time interval me repeat hota hai.
(c) Periodic. H₂ molecule ka rotation fixed angular speed se hota hai, isliye equal time intervals me same configuration repeat hoti hai.
(d) Not periodic. Ek baar release hone ke baad arrow seedha aage badhta hai — ye ek single non-repetitive event hai.
13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not SHM?
(a) The rotation of Earth about its axis.
(b) Motion of an oscillating mercury column in a U-tube.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower-most point.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
(a) Periodic, but NOT SHM (na hi oscillatory hai) — uniform circular motion hai, koi to-and-fro nahi.
(b) SHM — mercury column ek side push hone par restoring force displacement ke proportional hoti hai (extra weight of unbalanced column), isliye SHM.
(c) SHM (small displacement ke liye) — ye simple pendulum jaisa case hai, restoring force mg sinθ ≈ mgθ, linear approximation SHM deti hai.
(d) Periodic, but generally NOT simple SHM — polyatomic molecule ki vibration alag-alag normal modes (frequencies) ka superposition hoti hai, isliye complex periodic motion hai, single SHM nahi.
13.3 Figure (textbook diagram) depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?
(Standard NCERT figure ke conventional reading ke hisaab se — apni copy me diagram se cross-check zaroor karo):
(a) Periodic, T = 2 s (not SHM — shape sinusoidal nahi hai).
(b) Periodic, T = 2 s, aur ye curve SHM bhi hai (sinusoidal shape).
(c) Not periodic — pattern shuru me repeat hota dikhta hai par fixed interval par identical repeat nahi karta.
(d) Periodic, T = 2 s (not SHM — waveform symmetric sinusoid nahi hai).
Key test: koi bhi curve periodic tab hi hai jab woh apne aap ko ek fixed time interval T baad exactly repeat kare.
13.4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give the period for each case of periodic motion (ω is any positive constant).
(a) sin ωt − cos ωt
(b) sin³ωt
(c) 3 cos(π/4 − 2ωt)
(d) cos ωt + cos 3ωt + cos 5ωt
(e) exp(−ω²t²)
(f) 1 + ωt + ω²t²
(a) sin ωt − cos ωt = √2 sin(ωt − π/4) → single sinusoid → SHM, T = 2π/ω.
(b) sin³ωt = ¾ sin ωt − ¼ sin 3ωt → do alag frequencies ka combination → periodic, not SHM. LCM period = 2π/ω.
(c) 3 cos(π/4 − 2ωt) = 3 cos(2ωt − π/4) (cosine even function) → single sinusoid → SHM, T = 2π/(2ω) = π/ω.
(d) cos ωt + cos 3ωt + cos 5ωt → teen frequencies ka sum → periodic, not SHM. Periods 2π/ω, 2π/3ω, 2π/5ω ka LCM = 2π/ω.
(e) exp(−ω²t²) → exponentially decaying Gaussian, kabhi repeat nahi karta → non-periodic.
(f) 1 + ωt + ω²t² → t badhne ke saath monotonically badhta hai, kabhi repeat nahi hota → non-periodic.
13.5 A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A.
Mid-point O ko origin lo, A = −5 cm, B = +5 cm.
(a) At A (extreme, x = −5): v = 0, a = positive (mean ki taraf, i.e. B ki taraf), F = positive.
(b) At B (extreme, x = +5): v = 0, a = negative (mean ki taraf, i.e. A ki taraf), F = negative.
(c) Mid-point, going towards A (x = 0): v = negative, a = 0, F = 0 (mean position pe restoring force zero hoti hai).
(d) 2 cm from B towards A (x = +3, moving towards A): v = negative, a = negative (x positive hai isliye restoring acceleration negative), F = negative.
(e) 3 cm from A towards B (x = −2, moving towards B): v = positive, a = positive (x negative hai isliye a positive), F = positive.
(f) 4 cm from B towards A (x = +1, moving towards A): v = negative, a = negative, F = negative.
13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = −200x²
(c) a = −10x
(d) a = 100x³
SHM ki condition: a = −ω²x, yaani a, x ke directly (linearly) proportional ho aur opposite sign me ho.
(a) a = +0.7x → sign galat hai (positive proportional, restoring nahi hai) → SHM nahi.
(b) a = −200x² → x² ke proportional hai (non-linear) → SHM nahi.
(c) a = −10x → linear aur negative sign → yehi SHM hai. Yahan ω² = 10.
(d) a = 100x³ → x³ ke proportional aur sign bhi galat → SHM nahi.
13.7 The motion of a particle executing SHM is described by the displacement function x(t) = A cos(ωt + φ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s⁻¹. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions?
Cosine form: x(t) = A cos(ωt+φ) → v(t) = −Aω sin(ωt+φ)
t=0: A cos φ = 1 ...(i)
−Aω sin φ = ω ⟹ A sin φ = −1 ...(ii)
(i)² + (ii)²: A² = 1² + (−1)² = 2 ⟹ A = √2 cm
tan φ = (−1)/1 = −1, cos φ positive aur sin φ negative (4th quadrant) ⟹ φ = −π/4 rad
Sine form: x = B sin(ωt+α) → v = Bω cos(ωt+α)
t=0: B sin α = 1, Bω cos α = ω ⟹ B cos α = 1
B² = 1+1 = 2 ⟹ B = √2 cm; tan α = 1, dono positive (1st quadrant) ⟹ α = π/4 rad
13.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Spring constant:
k = (Force range)/(extension range) = (50 × 9.8 N)/(0.20 m) = 2450 N/m
Period formula se mass nikalo:
T = 2π√(m/k) ⟹ m = kT²/4π² = (2450 × 0.6²)/(4π²) = 882/39.48 ≈ 22.35 kg
Weight = mg = 22.35 × 9.8 ≈ 219 N
13.9 A spring having a spring constant 1200 N/m is mounted on a horizontal table. A mass of 3 kg is attached to the free end of the spring, pulled sideways to a distance of 2.0 cm, and released. Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
k = 1200 N/m, m = 3 kg, A = 0.02 m
ω = √(k/m) = √(1200/3) = √400 = 20 rad/s
(i) Frequency:
f = ω/2π = 20/(2π) ≈ 3.18 Hz
(ii) Maximum acceleration:
a_max = ω²A = 400 × 0.02 = 8 m/s²
(iii) Maximum speed:
v_max = ωA = 20 × 0.02 = 0.4 m/s
13.10 In Question 13.9, take the position of mass when the spring is unstretched as x = 0, and left-to-right as positive direction. Give x(t) if at t = 0 the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions differ — in frequency, amplitude, or initial phase?
ω = 20 rad/s, A = 2 cm (from Q13.9) — teeno cases me same rehte hain.
(a) Mean position se shuru (moving in +ve direction):
x(t) = 2 sin(20t) cm
(b) Maximum stretched (+ve extreme) se shuru:
x(t) = 2 cos(20t) cm
(c) Maximum compressed (−ve extreme) se shuru:
x(t) = −2 cos(20t) cm
Difference: Teeno functions ka amplitude (2 cm) aur angular frequency (20 rad/s ⟹ same T) same hai — sirf initial phase alag hai (0, aur π ka farak).
13.11 Figures correspond to two circular motions. The radius, period of revolution, initial position, and sense of revolution (clockwise/anticlockwise) diagram me diye gaye hain. Obtain the SHM of the x-projection of the radius vector of the revolving particle P in each case.
(Standard NCERT figure ke conventional data ke hisaab se — apni copy ke exact figure se labels verify kar lena.)
Case (x): radius = 3 cm, T = 2 s, particle anticlockwise, initial position OP 45° pe.
ω = 2π/T = π rad/s ⟹ x(t) = 3 cos(πt + π/4) cm
Case (y): radius = 2 m, T = 4 s, particle clockwise, initial position negative x-axis pe.
ω = 2π/T = π/2 rad/s ⟹ x(t) = −2 cos(πt/2) m
Method: Reference-circle approach me x-projection of a particle moving on a circle with angular speed ω, radius R, starting phase θ₀, hamesha x(t) = R cos(ωt + θ₀) (ya sin, initial position ke hisaab se) hoti hai — yahi SHM ki definition hai.
13.12 Plot the corresponding reference circle for each of the following SHMs. Indicate the initial (t=0) position, radius, and angular speed of the rotating particle:
(x) x = −2 sin(3t + π/3)
(y) x = cos(π/6 − t)
(z) x = 3 sin(2πt + π/4)
(x) −2 sin(3t+π/3) = 2 cos(3t + π/3 + π/2) = 2 cos(3t + 5π/6)
Radius = 2 units, ω = 3 rad/s, initial phase = 5π/6 rad (150°) — anticlockwise rotation.
(y) cos(π/6 − t) = cos(t − π/6) (cosine even function hai)
Radius = 1 unit, ω = 1 rad/s, initial phase = −π/6 rad (or π/6, clockwise sense se treat karo).
(z) 3 sin(2πt + π/4) = 3 cos(2πt + π/4 − π/2) = 3 cos(2πt − π/4)
Radius = 3 units, ω = 2π rad/s (T=1 s), initial phase = −π/4 rad (−45°) — anticlockwise rotation.
13.13 (a) A spring of force constant k, clamped at one end, mass m at free end, force F applied stretches it. In another setup, the same spring (both ends free) has mass m attached at each end, and each end pulled by force F. What is the maximum extension of the spring in the two cases? (b) If the masses are released, what is the period of oscillation in each case?
(a) Maximum extension: Dono cases me Newton's third law se spring ke andar tension F hi hai (case (b) me bhi har end pe F hi net force hai), isliye:
Extension = F/k (dono cases me SAME)
(b) Period:
Case (a) — one end fixed:
T₁ = 2π√(m/k)
Case (b) — dono ends free, do masses m: Ye reduced-mass problem hai, μ = m·m/(m+m) = m/2, spring constant same k:
T₂ = 2π√(μ/k) = 2π√(m/2k) = T₁/√2
13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with SHM with an angular frequency of 200 rad/min, what is its maximum speed?
Stroke = 1.0 m ⟹ Amplitude A = 0.5 m
ω = 200 rad/min = 200/60 rad/s ≈ 3.33 rad/s
v_max = ωA = 3.33 × 0.5 ≈ 1.67 m/s
13.15 The acceleration due to gravity on the surface of the moon is 1.7 m/s². What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of Earth is 3.5 s? (g_earth = 9.8 m/s²)
l constant rehti hai, T = 2π√(l/g) ⟹ T ∝ 1/√g
T_moon/T_earth = √(g_earth/g_moon) = √(9.8/1.7) = √5.76 = 2.4
T_moon = 3.5 × 2.4 = 8.4 s
13.16 Answer the following:
(a) T = 2π√(m/k) me time period force constant k aur mass m par depend karta hai — simple pendulum bhi approx SHM karta hai, phir T mass-independent kyun hai?
(b) Large angle oscillations me T, 2π√(l/g) se zyada kyun hoti hai?
(c) Wristwatch pehne aadmi tower se free-fall me girta hai — kya watch sahi time deta rahega?
(d) Ek freely-falling cabin me lagi simple pendulum ki oscillation frequency kya hogi?
(a) Simple pendulum ke liye restoring force = mg sinθ ≈ mgθ. Isse effective 'spring constant' k = mg/l khud mass m ke proportional hai. Isliye:
T = 2π√(m/k) = 2π√(m/(mg/l)) = 2π√(l/g)
Mass numerator aur denominator dono me cancel ho jaata hai — isliye T mass-independent hai.
(b) Bade angles pe sinθ < θ hoti hai, isliye actual restoring force (mg sinθ) idealized SHM force (mgθ) se kam hoti hai. Kamzor restoring force ka matlab hai particle ko mean position tak wapas aane me zyada time lagta hai — isliye T, 2π√(l/g) se badh jaata hai.
(c) Haan, wristwatch (spring/quartz-based mechanism) sahi time dega — kyunki iski working gravity par depend nahi karti (ye pendulum clock nahi hai). Sirf pendulum-based clocks free-fall me galat time denge.
(d) Free-fall me effective gravity g_eff = 0 (weightlessness). Chunki restoring force ∝ g_eff, isliye koi restoring force nahi bachta — pendulum oscillate hi nahi karega, frequency = 0.
13.17 A simple pendulum of length l and bob mass M is suspended in a car moving on a circular track of radius R with uniform speed v. If the pendulum makes small oscillations in a radial direction, what will be its time period?
Car me do accelerations act kar rahi hain: gravity g (vertically down) aur centripetal acceleration v²/R (horizontally, radially). Bob par effective gravity in dono ka resultant hogi:
g_eff = √(g² + (v²/R)²)
T = 2π√(l/g_eff) = 2π√[ l / √(g² + v⁴/R²) ]
13.18 Cylindrical piece of cork of base area A and height h, density ρ, floats in a liquid of density ρ_l. Cork depressed slightly and released — show it executes SHM with T = 2π√(hρ/ρ_l g). (Ignore viscosity/damping.)
Equilibrium me: Weight of cork = Buoyant force
ρAhg = ρ_l A h' g (h' = submerged height)
Cork ko x niche push karo — extra buoyant force upward:
F_restoring = −ρ_l A g x
Newton's second law (mass of cork m = ρAh):
ρAh · (d²x/dt²) = −ρ_l A g x
d²x/dt² = −(ρ_l g/ρh) x ⟹ ω² = ρ_l g/(ρh)
T = 2π/ω = 2π√(hρ/ρ_l g) — Hence proved.
13.19 One end of a U-tube containing mercury is connected to a suction pump, the other end to atmosphere. A small pressure difference is maintained. Show that when the suction pump is removed, the mercury column executes SHM.
Total mercury length l, density ρ, cross-section area A. Ek side x displace hone par doosri side bhi x rise karti hai, level difference = 2x, ye ek unbalanced weight create karta hai:
F_restoring = −(A · 2x · ρ · g) = −2Aρg·x
Total mass of mercury = Alρ:
(Alρ)(d²x/dt²) = −2Aρg x ⟹ d²x/dt² = −(2g/l) x
ω² = 2g/l ⟹ T = 2π√(l/2g) — SHM proved.
13.20 An air chamber of volume V has a neck of cross-section a into which a ball of mass m just fits, frictionless. Show that when pushed down slightly and released, the ball executes SHM. Find the period (pressure-volume variations are isothermal, chamber pressure = atmospheric P).
Ball ko x niche push karo → volume decrease dV = ax. Isothermal (Boyle's Law): PV = constant ⟹ P dV + V dP = 0
dP = −P(dV)/V = −P(ax)/V
Ye extra pressure ball par ek restoring force lagati hai:
F = −dP · a = −(Pa²/V) x
Newton's law: m(d²x/dt²) = −(Pa²/V)x
ω² = Pa²/(mV) ⟹ T = 2π√(mV/Pa²)
13.21 You are riding an automobile of mass 3000 kg. The suspension sags 15 cm under the full weight. Amplitude of oscillation decreases by 50% during one complete oscillation. Estimate (a) spring constant k, and (b) damping constant b for one wheel's suspension (each wheel supports 750 kg).
Mass per wheel m = 750 kg, sag x = 0.15 m
(a) Spring constant:
k = mg/x = (750 × 9.8)/0.15 ≈ 4.9 × 10⁴ N/m
(b) Damping constant: Natural period (damping halka maano):
T = 2π√(m/k) = 2π√(750/49000) ≈ 0.777 s
Amplitude decay: A(T) = A₀ e^(−bT/2m). 50% decrease ⟹ A(T)/A₀ = 0.5:
bT/2m = ln 2 ⟹ b = 2m ln2 / T = (2 × 750 × 0.693)/0.777 ≈ 1.33 × 10³ kg/s
13.22 (Additional Exercise) Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
x = A sin ωt ⟹ v = Aω cos ωt
KE = ½mv² = ½mA²ω² cos²ωt
PE = ½kx² = ½mω²A² sin²ωt (chunki k = mω²)
Time-average of sin²ωt aur cos²ωt dono over a full period = ½:
⟨KE⟩ = ¼mA²ω² = ⟨PE⟩
Dono equal hain aur total energy E = ½mω²A² ka aadha (½E) hote hain — Hence proved.
13.23 (Additional Exercise) A circular disc of mass 10 kg is suspended by a wire attached at its centre. The wire is twisted and released. Time period of torsional oscillation = 1.5 s, radius of disc = 15 cm. Find the torsional spring constant α (J = −αθ).
Moment of inertia of disc: I = ½MR² = ½ × 10 × (0.15)² = 0.1125 kg·m²
Torsional SHM: T = 2π√(I/α)
α = 4π²I/T² = (4π² × 0.1125)/(1.5)² = 4.441/2.25 ≈ 1.97 N·m/rad
13.24 (Additional Exercise) A body describes SHM with amplitude 5 cm and period 0.2 s. Find the acceleration and velocity when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
A = 0.05 m, T = 0.2 s ⟹ ω = 2π/T = 10π ≈ 31.4 rad/s
(a) x = 5 cm (extreme position):
v = 0 (extreme pe velocity zero); a = −ω²x = −(31.4)² × 0.05 ≈ −49.3 m/s² (max acceleration)
(b) x = 3 cm:
v = ω√(A²−x²) = 31.4 × √(0.0025−0.0009) = 31.4 × 0.04 ≈ 1.26 m/s
a = −ω²x = −986.96 × 0.03 ≈ −29.6 m/s²
(c) x = 0 (mean position):
v = ωA = 31.4 × 0.05 ≈ 1.57 m/s (max velocity); a = 0
13.25 (Additional Exercise) A mass attached to a spring oscillates with angular velocity ω in a horizontal plane without friction/damping. It is pulled to a distance x₀ and pushed towards the centre with velocity v₀ at t=0. Determine the amplitude of the resulting oscillations in terms of ω, x₀ and v₀.
Let x(t) = A sin(ωt + φ), v(t) = Aω cos(ωt + φ)
At t = 0: x = x₀ (pulled out), velocity pushed towards centre ⟹ v(0) = −v₀:
A sin φ = x₀ ...(i)
Aω cos φ = −v₀ ⟹ A cos φ = −v₀/ω ...(ii)
(i)² + (ii)² se:
A² = x₀² + v₀²/ω²
A = √(x₀² + v₀²/ω²)
Important Equations — Ek Nazar Me
| Quantity / Rule | Formula |
|---|---|
| Frequency – Period relation | f = 1/T |
| Angular frequency | ω = 2π/T = 2πf |
| SHM displacement (general) | x(t) = A cos(ωt + φ) or A sin(ωt + φ) |
| Velocity in SHM | v = −Aω sin(ωt+φ) = ω√(A²−x²) |
| Acceleration in SHM | a = −ω²x (defining condition of SHM) |
| Force law of SHM | F = −kx |
| Angular frequency — spring-mass | ω = √(k/m) |
| Time period — spring-mass | T = 2π√(m/k) |
| Time period — simple pendulum | T = 2π√(l/g) (mass-independent, small θ) |
| Kinetic energy in SHM | KE = ½mω²(A²−x²) |
| Potential energy in SHM | PE = ½mω²x² = ½kx² |
| Total energy in SHM | E = ½mω²A² = ½kA² (constant, all positions) |
| Springs in series | 1/k = 1/k₁ + 1/k₂ |
| Springs in parallel | k = k₁ + k₂ |
| Damped oscillation — amplitude decay | A(t) = A₀ e−bt/2m |
| Damped oscillation — angular frequency | ω' = √(k/m − b²/4m²) |
| Resonance condition (forced oscillation) | ωdriving = ωnatural ⟹ amplitude maximum |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- SHM = periodic motion samajh lena — saari SHM periodic hoti hai, lekin har periodic motion SHM nahi hoti (jaise Earth ka rotation periodic hai but SHM nahi). Check karo restoring force linear (F=−kx) hai ya nahi.
- a = −ω²x me negative sign bhool jana — acceleration hamesha mean position ki taraf hoti hai, displacement ke opposite direction me. Sign chhod dene se force/acceleration ka direction galat aa jaata hai.
- Simple pendulum ke T = 2π√(l/g) me mass daal dena — pendulum ka time period mass-independent hai (mass numerator-denominator me cancel ho jaata hai), sirf length aur g par depend karta hai.
- Phase angle calculate karte waqt radians aur degrees mix kar dena — ω hamesha rad/s me hoti hai, isliye phase φ bhi radians me hi rakho, calculator degree mode me na rakho.
- Spring ko cut karne par k same samajh lena — spring ko n equal parts me kaatne se har part ki spring constant k' = nk ho jaati hai (length kam hone se k badhta hai), yahi Q13.13 jaisi problems me galti hoti hai.
- Damped oscillation me amplitude decay ko linear soch lena — asal me ye exponential decay hai: A(t) = A₀e^(−bt/2m), linear nahi.
Board-Style Important Questions
- 1 mark: SHM ke liye acceleration aur displacement ke beech ka relation likho aur negative sign ka physical significance batao.
- 2 marks: Simple pendulum ka time period mass-independent kyun hota hai, dikhao.
- 3 marks: Ek particle SHM kar raha hai amplitude A aur angular frequency ω ke saath. Mean position se x distance par iski kinetic energy aur potential energy ke liye expressions derive karo, aur dikhao ki total energy constant rehti hai.
- 3 marks: Spring constant k wali spring se juda mass m block horizontal surface par SHM karta hai. Iska time period derive karo aur maximum velocity/acceleration ke expressions likho.
- 5 marks: Damped oscillation kya hoti hai? Iski amplitude, angular frequency, aur energy ke expressions likho, aur resonance ke saath iska comparison karo.
- 2 marks: Ek cylindrical cork liquid me float kar raha hai — dikhao ki chhota vertical displacement dene par ye SHM karta hai.
Aksar Poochhe Jaane Wale Sawaal
SHM aur periodic motion me kya fark hai?
Har SHM periodic hoti hai, lekin har periodic motion SHM nahi hoti. SHM ke liye zaroori hai ki restoring force/acceleration displacement ke directly proportional ho aur opposite direction me ho (F = −kx). Periodic motion sirf itna maangti hai ki motion fixed time interval baad repeat ho — jaise Earth ka rotation periodic hai lekin SHM nahi.
Simple pendulum ka time period mass par depend karta hai kya?
Nahi. T = 2π√(l/g) formula me mass hai hi nahi — kyunki restoring force (mg sinθ) aur inertia (mass m) dono hi mass ke proportional hote hain, isliye T = 2π√(m/k) me mass cancel ho jaata hai. Ye sirf pendulum ki length aur us jagah ki g par depend karta hai (small-angle approximation ke andar).
Class 11 Physics Chapter 13 Oscillations me total kitne exercise questions hain?
NCERT chapter ke end me total 25 exercise questions hain (13.1 se 13.25 tak, jisme 13.22–13.25 'Additional Exercises' hain) — inme conceptual (periodic/SHM identification, reference circle) aur numerical (spring, pendulum, damped oscillation) dono type ke questions hain.
Damped oscillation, forced oscillation aur resonance me kya difference hai?
Damped oscillation me resistive force (jaise friction) ki wajah se amplitude time ke saath exponentially decay hoti hai (A(t) = A₀e^(−bt/2m)). Forced oscillation me ek external periodic force system ko oscillate karne pe majboor karti hai, apni khud ki frequency chhod ke driving frequency pe. Jab driving frequency = system ki natural frequency ho jaati hai, tab amplitude maximum ho jaati hai — isi ko resonance kehte hain.
Class 11 Physics ke 2026-27 rationalised syllabus me total kitne chapters hain, aur Oscillations kaunsa chapter number hai?
2026-27 session ke rationalised syllabus (class 11 physics all chapters pdf 2026-27) me Class 11 Physics ab 14 chapters ka hai — purana standalone 'Physical World' chapter drop kar diya gaya hai, syllabus seedha Units and Measurements (Ch.1) se shuru hota hai. Is structure me Oscillations Chapter 13 hai, aur Waves (Ch.14) usse turant pehle nahi, baad me aata hai.
Class 11 Physics chapter-wise weightage me Oscillations ka kitna weightage hota hai CBSE board exam mein?
Is spec me kisi bhi exact marks-weightage number ka fabrication nahi kiya gaya hai kyunki official CBSE marking scheme har session update ho sakti hai. Class 11 physics chapter wise weightage cbse ke liye apne current academic session ke official CBSE curriculum document/sample paper se hi confirm karna sahi rahega — yahan sirf ye kaha ja sakta hai ki Oscillations, numerical-heavy chapters ki category me aata hai jisme SHM ke energy/force/time-period type questions frequently practice ke liye important rehte hain.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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