Chapter: Sound — Propagation of Sound - Long Answer Questions & Answers for CBSE Class 9 Physics

Medium Level (Application & Explanation)

Q1. Why can sound not travel through a vacuum? Explain using the bell-jar vacuum experiment.

Answer:

Sound is a mechanical wave that needs a material medium (solid, liquid, or gas) to travel because it transfers energy by causing vibrations of particles in the medium.
In a vacuum, there are virtually no particles to vibrate, so the chain of energy transfer cannot continue and sound cannot propagate.
In the bell-jar experiment, a ringing alarm inside the jar becomes faint and then silent as air is pumped out. This demonstrates that as the number of particles decreases, the audible sound weakens and finally disappears when an almost complete vacuum is reached.
The experiment clearly shows the necessity of a medium for sound propagation.

Q2. Explain why sound travels fastest in solids, slower in liquids, and slowest in gases. Give examples.

Answer:

The speed of sound depends on how quickly vibrations pass between particles and on the stiffness (elasticity) and density of the medium.
In solids, particles are closely packed and bonded strongly, so vibrations transfer rapidly; e.g., sound in steel ≈ 5000 m/s.
In liquids, particles are less tightly packed than in solids, so transfer is slower than solids but faster than gases; e.g., sound in water ≈ 1500 m/s.
In gases, particles are far apart, so collisions are less frequent and transfer is slowest; e.g., sound in air ≈ 343 m/s (at room temperature).
Thus particle spacing and bonding explain the order: solid > liquid > gas.

Q3. How does temperature affect the speed of sound in air? Relate this to the delay between lightning and thunder.

Answer:

In gases like air, temperature affects the average speed of molecules. Higher temperature means molecules move faster, so they transmit vibrations quicker.
The empirical relation is roughly v ≈ 331 + 0.6T (m/s), where T is temperature in °C; thus sound speed increases with temperature.
During a thunderstorm, you see the lightning (light travels at 3×10^8 m/s) almost instantly but hear thunder later because sound is much slower.
The time delay between lightning and thunder helps estimate the distance to the lightning using the known speed of sound, accounting for current air temperature to improve accuracy.

Q4. Describe how an echo is produced and how to use echo to measure the distance of a reflecting surface.

Answer:

An echo is produced when sound waves travel to a distant reflecting surface (like a cliff or wall) and are reflected back to the listener. The reflected sound reaches the listener after a measurable time delay.
To use an echo for distance measurement, produce a sharp sound and record the time (t) until the echo returns. The sound travels twice the distance (to the reflector and back), so distance d = (v × t) / 2, where v is the speed of sound in that medium.
For accurate results, ensure the reflecting surface is large and smooth, use a loud sharp sound, and account for air temperature to get the correct v.

Q5. Explain what you observe when you tap one end of a metal rod while a friend places their ear at the other end. Relate this to longitudinal waves and particle vibration.

Answer:

When you tap one end of a metal rod, you create a mechanical disturbance that causes particles in the rod to vibrate. These vibrations travel as longitudinal waves (particles oscillate along the direction of wave travel).
Your friend hears the sound sooner and clearer through the rod than through the air because solids transmit vibrations faster and with less loss.
The rod’s rigidity and close particle spacing allow the energy to pass quickly from particle to particle.
This experiment shows that sound can travel through solids effectively and that mechanical waves rely on particle interactions, not on the motion of the particles over long distances.

High Complexity (Analytical & Scenario-Based)

Q6. Scenario: You put your head underwater in a pool and a friend calls you from above the surface. Explain why you can still hear them and why the sound may seem different.

Answer:

Sound from your friend travels through air, hits the water surface, and part of it transmits into water while part reflects away. Because water is denser, there is partial reflection and partial transmission at the boundary.
The transmitted sound travels faster in water (~1500 m/s) and reaches your ear; you can therefore hear the voice even though the source is above water.
The voice may sound muffled or lower-pitched because of impedance mismatch, surface reflection filtering high frequencies, and because your ear under water receives sound mainly as bone-conducted and fluid-conducted vibrations, altering timbre.
Thus boundary effects and medium differences explain the change in quality and clarity.

Q7. Design a simple school experiment to compare the speed of sound in air, water, and a solid rod. Describe procedure, expected results, and sources of error.

Answer:

Procedure: For air, measure time for a clap echo from a wall at known distance; compute v_air = (2d)/t. For water, use two underwater hydrophones or have one person produce a sharp sound at one end of a long water-filled tube and record time to reach the other end; compute v_water = d/t. For a solid rod, strike one end and measure time to hear via contact at the other end (or use a stopwatch and two observers); compute v_solid = d/t.
Expected results: v_solid > v_water > v_air (e.g., steel ~5000 m/s, water ~1500 m/s, air ~343 m/s).
Sources of error: reaction time in measurements, imprecise distance, reflections, temperature variations (affecting v), and imperfect coupling between transducers and materials. Control by repeating trials and using electronic timing where possible.

Q8. Scenario: Two astronauts are on opposite sides of a spacecraft hull in space. One bangs the hull and the other feels a vibration through their suit. Explain how the sound/vibration traveled and why the bang was felt but not heard through space.

Answer:

In space there is no air, so air-borne sound cannot travel between astronauts. However, the mechanical impact on the spacecraft hull creates elastic waves that travel through the metal as solid-borne vibrations.
These vibrations propagate through the structure and through contact points into the astronaut’s suit and body, which they feel as vibrations. Inside the suit, the small amount of air allows some air-borne sound locally, so the astronaut may also hear a muffled noise.
Thus the hull acts as a solid medium, enabling vibration transmission across space where air-borne sound cannot propagate.

Q9. Explain how soundproofing works in a room using the principles of propagation, and suggest materials and construction features that reduce transmission of sound.

Answer:

Soundproofing reduces transmission, reflection, and resonance of sound using three main strategies: absorption, mass, and decoupling.
Absorptive materials (foam, mineral wool) convert sound energy into heat by making the air and fibers vibrate; they reduce echoes and reverberation.
Adding mass (dense walls, double-glazed windows) lowers transmission because heavier barriers are harder to vibrate.
Decoupling (double walls, resilient mounts) prevents structural vibrations from crossing by creating gaps or flexible connections.
Use seals around doors and windows to stop leaks, and porous, thick materials for low-frequency absorption. Combining these methods yields effective soundproofing.

Q10. Scenario: During a storm you see a lightning flash and hear thunder after 5 seconds. Calculate the distance to the lightning and explain assumptions and limitations of this method.

Answer:

Using an average speed of sound in air v ≈ 343 m/s (at 20°C), the distance to the lightning is d = v × t = 343 × 5 = 1715 meters, or about 1.7 km.
This method assumes that the sound traveled in a straight line through uniform air and that the recorded time is accurate. In practice, temperature, humidity, wind, and atmospheric layers can change sound speed and path, causing small errors. Also human reaction time influences measurement if using a stopwatch. Despite limitations, this quick method gives a good estimate of lightning distance.