Oscillations

1. Periodic and Simple Harmonic Motion

Motion that repeats itself at regular intervals is periodic. SHM is a special case where restoring force is proportional to displacement and directed towards mean position.

Restoring Force: F = -k x (Hooke's Law definition).
Differential Equation: d²x/dt² + ω²x = 0.
Displacement: x(t) = A cos(ωt + φ).
([A: Amplitude, ω: Angular Frequency, φ: Initial Phase]).

Velocity & Acceleration

Velocity (v): v = -Aω sin(ωt + φ) = ±ω√(A² - x²).
- Max at Mean Position: v_max = Aω.
- Min (Zero) at Extremes.
Acceleration (a): a = -Aω² cos(ωt + φ) = -ω²x.
- Max at Extremes: a_max = Aω².
- Zero at Mean Position.

(Asked in NEET 2016, 2019, 2021)

2. Dynamics of SHM

Kinetic Energy (K): ½mv² = ½mω²(A² - x²).
Potential Energy (U): ½kx² = ½mω²x².
Total Energy (E): K + U = ½mω²A² = Constant.
(Frequency of K and U is 2f, where f is frequency of SHM).

Spring-Mass Systems

Time Period: T = 2π√(m/k).
Series Combination: 1/k_eq = 1/k₁ + 1/k₂.
Parallel Combination: k_eq = k₁ + k₂.

Simple Pendulum

T = 2π√(L/g).
Independent of mass of bob.
Second's Pendulum: T = 2 sec (Length ≈ 1 m).

(Asked in NEET 2015, 2018, 2020, 2022)

3. Damped & Forced Oscillations

Real-world oscillations usually decay due to resistive forces (Damping).

Damped Oscillation: Amplitude decreases exponentially. A(t) = A₀ e^-bt/2m.
Forced Oscillation: Body oscillates with frequency of external periodic force.
Resonance: When driving frequency equals natural frequency, amplitude becomes maximum.

(Asked in NEET 2016, 2017)

Numericals: Oscillations

Q1. A mass M attached to spring has period 2s. If mass is increased by 2kg, period becomes 3s. Find M.

Solution:
T ∝ √m.
T₁ / T₂ = √(M / (M+2)).
2 / 3 = √(M / (M+2)).
4 / 9 = M / (M+2) → 4M + 8 = 9M → 5M = 8.
M = 1.6 kg.

Q2. At what displacement is the Kinetic Energy equal to Potential Energy in SHM?

Solution:

K = U → ½mω²(A² - x²) = ½mω²x².
A² - x² = x² → 2x² = A².
x = ± A / √2.

Q3. Max velocity of particle in SHM is v₀ and max acceleration is a₀. Find Time Period.

Solution:
v₀ = Aω.
a₀ = Aω².
dividing: a₀ / v₀ = ω.
T = 2π / ω = 2π(v₀ / a₀).

Q4. Length of simple pendulum increased by 44%. Find percentage increase in Time Period.

Solution:
T ∝ √L.
L' = 1.44 L.
T' ∝ √(1.44 L) = 1.2 √L = 1.2 T.
Increase = (1.2 - 1) * 100 = 20%.

Q5. Two springs k₁ and k₂ are connected in series. Find effective spring constant.

Solution:
For series combination:
1/k_eq = 1/k₁ + 1/k₂.
k_eq = (k₁k₂) / (k₁ + k₂).

Q6. Particle executes SHM with eq x = 5 cos(2πt + π/4) meters. Find displacement at t=1s.

Solution:
At t = 1s, phase = 2π(1) + π/4 = 2π + π/4.
cos(2π + π/4) = cos(π/4) = 1/√2.
x = 5 (1/√2) = 5/√2 m.

Q7. Particle of mass 0.1kg executes SHM with amplitude 0.1m and frequency 10/π Hz. Find Total Energy.

Solution:
E = ½ m ω² A².
ω = 2πf = 2π(10/π) = 20 rad/s.
E = 0.5 × 0.1 × (20)² × (0.1)².
E = 0.5 × 0.1 × 400 × 0.01 = 0.2 J.

Q8. Two particles Execute SHM x₁ = A sin(ωt) and x₂ = A cos(ωt). Phase difference?

Solution:
Convert cos to sin: cos(ωt) = sin(ωt + π/2).
Phases are ωt and ωt + π/2.
Difference Δφ = π/2.

Q9. Amplitude A=4cm. At what displacement is velocity half of maximum velocity?

Solution:
v = ω√(A²-x²) and v_max = Aω.
Given v = v_max/2 → √(A²-x²) = A/2.
Square both sides: A²-x² = A²/4.
x² = 3A²/4 → x = √3 A / 2.

Q10. Spring is cut into two equal halves. One half is used to suspend mass M. New Time Period? (Old was T).

Solution:
Cut in half → k' = 2k (stiffness doubles).
T = 2π√(m/k).
T' = 2π√(m/2k) = T / √2.
New Period = T / √2.

Important Formulae

1. Kinematics of SHM

Displacement:

x = A cos(ωt + φ)

Velocity:

v = ±ω √(A² - x²)

Acceleration:

a = - ω² x

2. Energy in SHM

Kinetic Energy:

K = ½ m ω² (A² - x²)

Potential Energy:

U = ½ m ω² x²

Total Energy:

E = ½ m ω² A² (Constant)

3. Time Period Systems

Spring-Mass System:

T = 2π √(m / k)

Simple Pendulum:

T = 2π √(L / g)

Springs in Parallel:

k_eq = k₁ + k₂

20 NEET Golden Facts

1. Defining SHM: Acceleration must be proportional to displacement and opposite to it. (a ∝ -x).
2. Mean Position: Point of stable equilibrium where Net Force = 0.
3. Extremes: Velocity is zero, Acceleration is maximum.
4. Phase (ωt+φ): State of oscillation (position and direction) at time t.
5. Graph v vs x: Ellipse. (Circle if scales adjusted).
6. Graph a vs x: Straight line with negative slope passing through origin.
7. Average Velocity: Over one complete cycle = 0.
8. Average Energy: Avg. Kinetic = Avg. Potential = Total/2 = ¼mω²A².
9. Pendulum Length: Distance from point of suspension to COM of bob.
10. Infinite Length Pendulum: T is not infinity, it is 84.6 minutes (Radius of Earth limit).
11. Cutting Spring: If spring of length L is cut into n equal parts, constant becomes nk.
12. Lift Motion: Pendulum in accelerating lift UP → g_eff = g+a, T decreases.
13. Free Fall: Pendulum in satellite/free fall → g_eff = 0, T = ∞ (No oscillation).
14. U-Tube Liquid: Oscillates with T = 2π√(h/g) where h is length of liquid column.
15. Two Particle System: Reduced mass μ = m₁m₂/(m₁+m₂) used in formula.
16. Phase Diff (v & x): Velocity leads Displacement by π/2.
17. Phase Diff (a & x): Acceleration leads Displacement by π (Out of phase).
18. Superposition: Two SHMs same freq, perpendicular → form Ellipse/Circle/Line (Lissajous figures).
19. Resonance sharpness: Depends on damping. Less damping → Sharper resonance peak.
20. Energy Freq: If SHM freq is f, KE/PE freq is 2f, Total Energy freq is 0.

📱 Practice MCQs for this topic inside our App