Q4: Sound Wave Analysis in Monoatomic Solid Lattice
Question:
(a) If the velocity of sound in a solid is 1 × 10³ m/s, determine the frequency of a sound wave with wavelength 10 Å (angstroms), assuming a one-dimensional monoatomic lattice.
(b) For a one-dimensional monoatomic lattice of lead (Pb), calculate the angular frequency of vibration (ω) and the maximum amplitude at room temperature (T = 300 K). Given:
- Young’s modulus: Y = 15 × 10⁹ N/m²
- Lattice parameter: a = 4.920 Å
- Atomic mass of lead: 207 amu (1 amu = 1.66 × 10⁻²⁷ kg)
Answer:
(a) Frequency of the Sound Wave
We use the wave relation:
f = v / λ
- v = 1 × 10³ m/s
- λ = 10 Å = 10 × 10⁻¹⁰ m
f = (1 × 10³) / (10 × 10⁻¹⁰) = 1 × 10¹⁴ Hz
✔️ Frequency of the wave is: 1 × 10¹⁴ Hz
(b) Angular Frequency and Maximum Amplitude
Step 1: Calculate Angular Frequency (ω)
In a monoatomic lattice, the angular frequency is given by:
ω = √(4Y / Ma²)
- Y = 15 × 10⁹ N/m²
- a = 4.920 Å = 4.920 × 10⁻¹⁰ m
- M = 207 amu = 207 × 1.66 × 10⁻²⁷ kg = 3.4362 × 10⁻²⁵ kg
Substitute into the formula:
ω = √[(4 × 15 × 10⁹) / (3.4362 × 10⁻²⁵ × (4.920 × 10⁻¹⁰)²)]
ω = √[6.0 × 10¹⁰ / 8.320 × 10⁻⁴⁴] = √(7.21 × 10⁵³) ≈ 8.49 × 10²⁶ rad/s
Step 2: Calculate Maximum Amplitude (A) at T = 300 K
Using the thermal energy relation:
½ Mω²A² = kBT ⟹ A = √(2kBT / Mω²)
- kB = 1.38 × 10⁻²³ J/K
- T = 300 K
- M = 3.4362 × 10⁻²⁵ kg
- ω = 8.49 × 10²⁶ rad/s
A = √[(2 × 1.38 × 10⁻²³ × 300) / (3.4362 × 10⁻²⁵ × (8.49 × 10²⁶)²)]
First calculate denominator:
(8.49 × 10²⁶)² = 7.21 × 10⁵³
Now calculate amplitude:
A = √(8.28 × 10⁻²¹ / 2.477 × 10²⁹) = √(3.34 × 10⁻⁵⁰) = 5.78 × 10⁻²⁵ m
✅ Final Answers Summary:
- Frequency of the sound wave (f): 1 × 10¹⁴ Hz
- Angular frequency (ω): 8.49 × 10²⁶ rad/s
- Maximum amplitude (A): 5.78 × 10⁻²⁵ m