PHYSICS 2 – FINAL TERM STUDY PACK
(Based on Lecture 4, 5, and 6 slides)
Lecture 4 & 5 – Mechanical Waves (Waves on a String) [Slide 20]
Key Theory
1) Wave Types: - Mechanical waves: need a medium; examples: sound, water waves,
seismic. - Electromagnetic waves: travel in vacuum at 3.0×10^8 m/s; examples: light, radio.
- Matter waves: wave nature of particles like electrons and protons.
2) Transverse vs Longitudinal: - Transverse: particles vibrate perpendicular to propagation.
- Longitudinal: particles vibrate parallel to propagation.
3) Wave Equation: y(x, t) = A sin(kx ± ωt + φ) A=amplitude, k=2π/λ, ω=2π/T=2π f, φ=phase
constant. Rule: +ωt → wave moves in +x, −ωt → −x.
4) Centripetal acceleration: a = v^2 / R.
5) Wave speed on string: v = sqrt(T/µ). Depends only on tension T and µ, not frequency.
Derivations (Lecture 4 & 5)
• Wave number: Over one wavelength λ, phase increases 2π → k=2π/λ. • Angular
frequency: Over one period T, phase increases 2π → ω=2π/T, f=1/T. • Wave speed
relation: From kx±ωt=constant → v=ω/k=λ f. • Newton’s 2nd law: For string element µ∆x,
radial force≈Tθ, acceleration=v²/R → v=√(T/µ).
Solved Problems (Lecture 4 & 5)
Q6. Wave y(x,t)=(6.0 mm) sin[(kx+600 t)+φ]. At t=0, x=0, y=0.01273 m. Solution: - At t=0,
x=0 → y= A sinφ = 0.01273 m - Given A=0.006 m (6.0 mm). - So amplitude y_m=0.01273
m=1.27 cm. Answer: 1.27 cm ✔
Q14. Wave y=(2.0 m) sin[(20 m^-1)x − (600 s^-1)t], T=15 N. Step 1: k=20 rad/m →
λ=2π/20=0.314 m. Step 2: ω=600 rad/s → f=600/2π≈95.5 Hz. Step 3: v=ω/k=600/20=30
m/s. Step 4: v²=900, µ=T/v²=15/900=0.0167 kg/m=16.7 g/m. Answer: v=30 m/s, µ≈16.7
g/m ✔
MCQs (Lecture 4 & 5)
1) Which type of wave requires a medium? → B (Mechanical) 2) Wave speed on a
stretched string depends on? → C (Tension & µ) 3) Wave number k=? → B (2π/λ) 4) In y=A
sin(kx+ωt+φ), wave moves in +x direction → Answer B
Lecture 6 – Superposition & Standing Waves [Slide 21]
Key Theory
1) Superposition principle: y′(x,t)=y1+y2. Displacements add algebraically. 2) Interference:
Two equal waves same direction → A′=2A cos(∆φ/2). 3) Standing waves: y′=2A sin(kx)
cos(ωt). Nodes: x=n λ/2 Antinodes: x=(2n+1) λ/4 Distance between adjacent nodes=λ/2
Derivation (Lecture 6)
Two waves moving opposite directions: y1=A sin(kx−ωt), y2=A sin(kx+ωt). Add them: y′=2A
sin(kx) cos(ωt). This is the standing wave. Amplitude at position x=2A sin(kx).
Solved Problems (Lecture 6)
Q32. Two identical waves same direction. Resultant amplitude=1.5A. A′=2A cos(∆φ/2). 2
cos(∆φ/2)=1.5 → cos(∆φ/2)=0.75. ∆φ/2=cos^-1(0.75)=41.41° So ∆φ=82.8°=1.45 rad=0.231
λ Answer: (a)82.8°, (b)1.45 rad, (c)0.231 λ ✔
Q53. y′=(0.50 cm) sin[(π/3 cm^-1)x] cos[(40π s^-1)t]. Step 1: 2y_m=0.0050 m →
y_m=0.0025 m Step 2: k=π/3 cm^-1=π/0.03=104.72 rad/m → λ=0.06 m Step 3:
ω=40π=125.66 rad/s (a) Amplitude=0.0025 m (b) v=ω/k=125.66/104.72≈1.20 m/s (c)
Distance between nodes=0.03 m (d) v_particle=−2Aω sin(kx) sin(ωt). At x=0.015 m →
kx=0.0157 rad (sin≈0.0157). At t=9/8 s → ωt≈141.37 rad≈45π → sin=0 → v_p=0 Answer:
(a)0.0025 m, (b)1.20 m/s, (c)0.03 m, (d)0 m/s ✔
Q56. y(x,t)=0.040 (sin 5πx)(cos 40πt). Nodes where sin(5πx)=0. 5πx=nπ → x=n/5 m. (i)0 m,
(ii)0.20 m, (iii)0.40 m. Answer: Nodes at 0 m, 0.20 m, 0.40 m ✔
MCQs (Lecture 6)
1) Superposition principle → Displacements add algebraically. Answer C 2) Resultant
amplitude = 2A cos(∆φ/2). Answer B 3) Node positions = n λ/2. Answer B 4) Antinodes =
(2n+1) λ/4. Answer C
End of Study Pack
