MP Board · Class 9 · Science · Sound(a) Differentiate between longitudinal waves and transverse waves with clear examples for each. (b) Describe the key characteristics of a sound wave: Amplitude, Wavelength, Frequency, and Time Period. State the mathematical relationship between wave speed ($v$), frequency ($\nu$), and wavelength ($\lambda$). (c) A sound wave travels at a speed of $340\text{ m/s}$ in air. If its wavelength is $1.7\text{ cm}$, calculate its frequency. Will this sound be audible to human ears? Give reasons.
(a) Difference Between Longitudinal and Transverse Waves
| Feature | Longitudinal Wave | Transverse Wave |
|---|---|---|
| Particle Displacement | Particles of the medium vibrate parallel to the direction of wave propagation. | Particles of the medium vibrate perpendicular to the direction of wave propagation. |
| Form of Propagation | Travels in the form of Compressions ($C$) and Rarefactions ($R$). | Travels in the form of Crests and Troughs. |
| Medium Requirement | Requires a material medium (solids, liquids, or gases) to propagate. | Can travel in solids, on liquid surfaces, or through vacuum (for electromagnetic waves). |
| Examples | Sound waves in air, waves in a compressed spring (slinky). | Waves on a plucked guitar string, light waves, water ripples. |
(b) Characteristics of a Sound Wave
-
Amplitude ($A$):
The magnitude of the maximum displacement of the particles of the medium from their mean position. It determines the loudness of the sound (larger amplitude = louder sound). -
Wavelength ($\lambda$):
The distance between two consecutive compressions or two consecutive rarefactions. Measured in metres ($\text{m}$). -
Frequency ($\nu$ or $f$):
The number of complete wave cycles or oscillations produced per unit time (per second). Measured in Hertz ($\text{Hz}$). -
Time Period ($T$):
The time taken by the wave to complete one full oscillation. Relation: $T = \frac{1}{\nu}$.
Mathematical Relation: $$\text{Wave Speed } (v) = \frac{\text{Distance}}{\text{Time}} = \frac{\lambda}{T} = \nu \cdot \lambda$$ $$\mathbf{v = \nu \lambda}$$
(c) Numerical Solution
Given:
- Speed of sound, $v = 340\text{ m/s}$
- Wavelength, $\lambda = 1.7\text{ cm} = \frac{1.7}{100}\text{ m} = 0.017\text{ m}$
Formula: $$v = \nu \times \lambda \implies \nu = \frac{v}{\lambda}$$
Calculation: $$\nu = \frac{340\text{ m/s}}{0.017\text{ m}} = 20000\text{ Hz} = 20\text{ kHz}$$
Audibility Conclusion:
- The frequency of this sound wave is $20,000\text{ Hz}$ ($20\text{ kHz}$).
- The normal human audible frequency range is between $20\text{ Hz}$ and $20,000\text{ Hz}$.
- Therefore, yes, this sound lies exactly at the upper limit of the human audible range and will be audible (especially to children and young individuals whose high-frequency hearing is intact).