NCERT Plus One (Class 11) Physics • Chapter 13

Oscillations

Periodic vs oscillatory motion, SHM & circular projection, v & a phase shifts, spring-mass, SHM energy & simple pendulum.

📄 Official NCERT Textbook PDF

Topic 13.2: Periodic vs Oscillatory Motion, Period & Frequency

Think of Earth spinning around the Sun once every 365 days — it repeats identically, yet it keeps going round and round without ever reversing back and forth. Now picture a bamboo Oonjal (swing) during Onam in Kottayam — it swings to and fro across one cosy resting middle spot! Every swing is periodic, yet NOT every periodic trip is oscillatory!
PERIODIC VS OSCILLATORYPeriodic: repeats every T (e.g. Earth orbit). Oscillatory: to-and-fro about a mean position!
PERIOD & FREQUENCYν = 1 / T (Hz = s⁻¹) • Angular freq ω = 2πν = 2π / T (rad/s)
FOURIER IDEAANY periodic function can be built by adding sine & cosine waves!
🧠 Memory Trick: All Oscillatory motions are Periodic (if undamped), but NOT all Periodic motions are Oscillatory (like uniform circular motion or a ceiling fan)!
💡 Simple Kid-Check: An oscillation is a "back-and-forth dance" around one favourite equilibrium point (like a guitar string, pendulum, or bouncing diving board). Time for ONE full dance is Period T; how many dances fit in 1 second is Frequency ν!
  • 1
    Time Period (T): Smallest time interval after which the motion repeats itself identically (unit: second s).
  • 2
    Frequency (ν): Number of complete repetitions per second, ν = 1/T (unit: Hertz Hz).
  • 3
    Displacement Variable: Need not always be distance! It can be angle θ of a swing, voltage V(t) in an AC plug, or pressure ΔP in sound!
Classic 1-mark Exam Question: Is uniform circular motion oscillatory? NO! It is periodic (repeats every T), NOT oscillatory, because there is no back-and-forth motion about an equilibrium position!

Topic 13.3 – 13.4: Simple Harmonic Motion (SHM) & Uniform Circular Projection

Watch a glowing Ferris wheel spinning steadily in the night at Marine Drive, Kochi — now look ONLY at the shadow of one cabin cast on the flat ground below! Even though the cabin circles at steady speed, its shadow slides smoothly back and forth — speeding through the middle and pausing gently at both ends. That exact shadow dance IS Simple Harmonic Motion (SHM)!
SHM EQUATIONx(t) = A cos(ωt + φ) (or A sin(ωt + φ))
KEY PARAMETERSA = Amplitude (max |x|), (ωt + φ) = Phase, φ = Initial Phase (Epoch)
UCM PROJECTIONProjection of Uniform Circular Motion on any diameter is pure SHM!
🧠 Memory Trick: Amplitude (A) = radius of the reference circle; ω = angular speed of rotation; φ = starting clock angle at t = 0!
💡 Simple Kid-Check: SHM is the simplest, smoothest kind of oscillation — shaped like a pure sine or cosine wave! Its displacement bounces symmetrically between −A and +A forever.
  • 1
    Phase (ωt + φ): Tells you BOTH where the particle is AND which way it is moving at time t! Adding 2π brings the particle back to the exact same state.
  • 2
    Initial Phase (φ): If you start at +A when t = 0, x = A cos(ωt) (φ = 0). If you start at mean x = 0 moving right, x = A sin(ωt)!
  • 3
    Superposition Trap Check: A single frequency like sin(ωt) + cos(ωt) = √2 sin(ωt + π/4) IS pure SHM! Adding two different frequencies like sin(ωt) + sin(2ωt) is periodic, NOT SHM!
Board Exam Favourite: Is y = sin(ωt) + cos(ωt) SHM? YES! Its amplitude is √(1² + 1²) = √2 (NOT 1 + 1 = 2!). Meanwhile, sin(ωt) + sin(3ωt) is periodic (period 2π/ω) but NOT SHM!

Topic 13.5: Velocity & Acceleration in SHM (Phase Differences)

Picture yourself on a swing at the very highest point — for one split second you freeze completely (v = 0), YET your stomach feels the strongest pull rushing back toward the centre (|a| = max)! Now swoop right through the bottom middle — you fly at top speed (|v| = max), while the pull back disappears (a = 0)!
VELOCITY IN SHMv = −Aω sin(ωt + φ) • v = ±ω √(A² − x²) • v_max = Aω
ACCELERATION IN SHMa = −ω² A cos(ωt + φ) = −ω² x • a_max = ω² A
PHASE LEAD RULESv leads x by π/2 (90°) • a leads x by π (180°)!
🧠 Memory Trick: At Mean (x = 0): |v| = MAX, a = 0! At Extreme (x = ±A): v = 0, |a| = MAX! Remember a = −ω² x is SHM's signature passport!
💡 Simple Kid-Check: Differentiate x = A cos(ωt + φ) once to get velocity, and differentiate a second time to get acceleration! Notice that acceleration always gets a minus sign (a = −ω²x) — meaning whenever you go right (+x), acceleration pulls you back left!
v_max = ω A (at mean x = 0)  |  a_max = ω² A (at extremes x = ±A)
Phase Gaps: x → v is +π/2  |  v → a is +π/2  |  x → a is +π
🧒 Quick Numerical: An SHM particle has amplitude A = 0.1 m and ω = 10 rad/s. Find max speed and max acceleration.
v_max = ωA = 10 × 0.1 = 1.0 m/s;   a_max = ω²A = 100 × 0.1 = 10 m/s²!
Graph Trap: What is the shape of the v vs x graph in SHM? From v² = ω²(A² − x²), dividing gives (x/A)² + (v/(Aω))² = 1 — an ELLIPSE! Meanwhile, the a vs x graph is a straight line with negative slope −ω²!

Topic 13.6: Force Law & Angular Frequency of Spring-Mass System

Hang a brass weight on a steel spring — pull it down slightly and let go! The harder you stretch it, the harder the spring tugs back toward equilibrium (F = −kx). Notice a giant spring-gun surprise: gravity pulls the hanging spring to a lower equilibrium spot, YET it does NOT change the up-and-down bounce time period at all!
SHM FORCE LAWF = −k x (Restoring force proportional & opposite to x!)
SPRING FREQUENCY & PERIODω = √(k / m)  ⇒  T = 2π √(m / k)
SPRINGS COMBINEDParallel: k_eq = k₁ + k₂ • Series: 1/k_eq = 1/k₁ + 1/k₂
🧠 Memory Trick: Heavy block (↑m) = Lazy, slower period (T ∝ √m); Stiff thick spring (↑k) = Snappy, shorter period (T ∝ 1/√k)!
💡 Simple Kid-Check: Compare Newton's Law F = ma = m(−ω²x) with Hooke's spring force F = −kx. Match the coefficients of x: mω² = k, which instantly gives ω = √(k/m) and T = 2π√(m/k)!
  • 1
    Restoring Condition: Any linear system obeying F = −kx (where k is force constant in N/m) executes pure SHM!
  • 2
    Vertical vs Horizontal Spring: Even if you take a vertical spring-mass to the Moon or inside a free-falling lift, its oscillation period T = 2π√(m/k) stays 100% UNCHANGED (independent of g)!
  • 3
    Cutting a Spring: Force constant scales inversely with length (k ∝ 1/L). Cut a spring in half → each half gets TWICE as stiff (k' = 2k)!
Classic Trap: Take a spring-mass oscillator AND a simple pendulum to the Moon (g_Moon = g/6). The pendulum slows down (T ∝ 1/√g), whereas the spring-mass period stays IDENTICAL because T = 2π√(m/k) has ZERO g in it!

Topic 13.7: Kinetic, Potential & Total Energy in SHM

Picture a skater gliding inside a smooth U-shaped bowl! At the top rim (±A), the skater stops for a beat — 100% stored Potential Energy, zero Kinetic. Swooping through the bottom centre (0), every bit turns into blazing Kinetic Energy! Halfway in energy, the two trade back and forth endlessly while total energy stays rock solid!
ENERGY FORMULASU = ½ k x² • K = ½ k (A² − x²) • E = ½ k A² = ½ m ω² A²
WHERE K = U ?At x = ± A / √2 (~0.707 A), Kinetic & Potential are both ½ E!
ENERGY FREQUENCYIf SHM frequency is ν, both K & U oscillate at 2ν!
🧠 Memory Trick: At x = A/2 (halfway in distance), U = ¼ E and K = ¾ E! Equal split (K = U) does NOT happen at A/2 — it happens at A / √2!
💡 Simple Kid-Check: Plot Potential Energy U = ½ k x² — it forms an upright U-shaped parabola! Plot Kinetic Energy K = E − U — it forms an upside-down arch! Add them at ANY position x and they always total the exact flat ceiling E = ½ k A²!
K(t) = ½ k A² sin²(ωt + φ)  |  U(t) = ½ k A² cos²(ωt + φ)
E_total = K + U = ½ k A² (sin² + cos²) = ½ k A² (Constant!)
🧒 Quick Check at x = A / 2:
U = ½ k (A/2)² = ¼ (½ k A²) = 25% of E  ⇒  K = E − U = 75% of E!
Two classic traps! (1) Doubling amplitude A makes Total Energy 4 TIMES larger (E ∝ A²). (2) In one complete oscillation of period T, Kinetic Energy hits maximum TWICE (at +v_max and −v_max), so energy frequency is 2ν!

Topic 13.8: The Simple Pendulum Time Period Derivation & Graph

Look at an old wooden grandfather clock ticking steadily in a heritage home in Fort Kochi! Galileo noticed a wild secret while watching a swinging church chandelier: whether the swing bob is light brass or heavy iron, as long as the angle stays small and the string length L stays fixed, every tick takes the exact same time!
PENDULUM PERIODT = 2π √(L / g) (Independent of bob mass m & small amplitude!)
SECONDS PENDULUMHas T = 2 s (1 sec each swing) ⇒ Length on Earth L ≈ 1 m (0.993 m)!
T² vs L GRAPHT² = (4π²/g) L → Straight line through origin with slope 4π²/g!
🧠 Memory Trick: "Tea equals 2π root L over g"! Wait — if a girl standing on a swing stands UP, her Centre of Mass rises → L shrinks → T shrinks (swings FASTER)!
💡 Simple Kid-Check: Gravity pulls the pendulum bob straight down with weight mg. Split mg along and perpendicular to the string: mg cos θ balances string tension, while the sideways component −mg sin θ pulls it back! For small angles (θ < 10°), sin θ ≈ θ (in radians), giving pure SHM!
  • 1
    Restoring Torque: Torque about pivot is τ = −L (mg sin θ) ≈ −mgL θ (for small θ).
  • 2
    Rotational Newton's Law: Since moment of inertia is I = mL² and τ = I α, we get mL² α = −mgL θ ⇒ α = −(g/L) θ.
  • 3
    Compare with α = −ω²θ: Bob mass m cancels cleanly! Thus ω = √(g/L) and T = 2π √(L / g)!
Four classic board traps on T = 2π√(L/g): (1) Changing bob mass does NOTHING to T. (2) Inside a freely falling lift, g_eff = 0 ⇒ T = ∞ (pendulum doesn't oscillate!). (3) In summer heat, metal wire expands → ↑L → ↑T → clock runs SLOW! (4) Measure L from pivot to centre of mass of bob (if hollow bob drains water, COM drops first then snaps back!).

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): "All oscillatory motions are periodic, but all periodic motions are not oscillatory." Justify with an example.

An oscillating pendulum repeats its path in equal time T (so every oscillatory motion is periodic). However, uniform circular motion (like Earth orbiting Sun) repeats regularly without moving to-and-fro about a mean position (periodic, NOT oscillatory).

Q2 (Kerala HSE March • 3 Marks): Derive expressions for Kinetic Energy, Potential Energy, and Total Energy of a particle executing SHM, and show that total mechanical energy is conserved.

For x = A cos(ωt + φ) and v = −Aω sin(ωt + φ):
• U = ½ k x² = ½ mω²A² cos²(ωt + φ)
• K = ½ m v² = ½ mω²A² sin²(ωt + φ)
• E = K + U = ½ mω²A² (sin² + cos²) = ½ mω²A² = ½ k A² (constant, independent of t and x).

Q3 (Kerala HSE Say • 3 Marks): Show that the oscillations of a simple pendulum are simple harmonic for small angular displacements, and derive T = 2π√(L/g).

Restoring torque τ = −mgL sinθ. For small θ, sinθ ≈ θ ⇒ τ ≈ −mgL θ.
Using τ = Iα = (mL²)α: mL²α = −mgL θ ⇒ α = −(g/L)θ.
Comparing with α = −ω²θ gives ω = √(g/L) and T = 2π√(L/g).

Q4 (Kerala HSE Model • 2 Marks): What is a seconds pendulum? Calculate its length on Earth where g = 9.8 m/s².

A seconds pendulum has a time period of exactly T = 2 s (taking 1 s per half-swing).
2 = 2π√(L/9.8) ⇒ L = 9.8 / π² ≈ 0.993 m (≈ 1 m).

Q5 (Kerala HSE March • 2 Marks): A body oscillates with SHM according to x = 5 cos(2πt + π/4) in SI units. At t = 1.5 s, calculate (a) displacement and (b) speed.

A = 5 m, ω = 2π rad/s. At t = 1.5 s, phase = 3π + π/4:
(a) x = 5 cos(3π + π/4) = −5 cos(π/4) = −3.535 m.
(b) |v| = |−Aω sin(3π + π/4)| = 5(2π)(1/√2) = 22.2 m/s.

Q6 (Kerala HSE Say • 2 Marks): A spring of force constant k is cut into two equal halves. What is the force constant of each half? If one half carries mass m, find the new period compared to original T.

Spring constant k ∝ 1/L. Halving length doubles stiffness: k' = 2k.
New period T' = 2π√(m / 2k) = T / √2.

Q7 (NCERT / DHSE Model • 2 Marks): What happens to the time period of (a) a simple pendulum and (b) a spring-mass system inside a satellite or freely falling lift?

(a) Effective gravity g_eff = g − a = 0, so pendulum period T = 2π√(L/g_eff) → ∞ (it does not oscillate).
(b) Spring period T = 2π√(m/k) is independent of g, so it remains unchanged.

Q8 (Kerala HSE March • 2 Marks): At what positions in SHM are (i) velocity maximum and zero, and (ii) acceleration maximum and zero?

(i) Velocity v = ±ω√(A² − x²) is maximum (ωA) at mean position (x = 0) and zero at extremes (x = ±A).
(ii) Acceleration a = −ω²x is zero at mean (x = 0) and maximum (ω²A) at extremes (x = ±A).