Chapter: Sound — Long Answer Questions (Class 9 CBSE)
Medium Level (Application & Explanation)
Q1. Explain how sound is produced when a guitar string is plucked and compare this with the sound produced when a drum is struck. Include how the surrounding medium participates in producing the sound you hear.
Answer:
When a guitar string is plucked, the string undergoes a rapid back-and-forth motion called vibration. These vibrations cause the nearby air particles to oscillate, creating regions of compression and rarefaction (pressure variations). These pressure variations travel as longitudinal sound waves through the air until they reach our ears. The guitar’s body amplifies the sound by transferring vibrations to larger air volumes.
In a drum, striking the membrane makes the membrane vibrate. The membrane pushes on the air more directly and with larger area, producing stronger pressure changes.
In both cases the medium (air) is essential: it carries the pressure variations from the vibrating object to the listener. Without a medium, the vibrations would not reach our ears and no sound would be heard. The nature of vibration (shape, frequency, amplitude) decides the pitch and loudness.
Q2. Why cannot sound travel through a vacuum? Describe a simple demonstration that shows this fact and explain the observations.
Answer:
Sound needs a medium (solid, liquid, or gas) because it travels by successive collisions between particles. In a vacuum there are no particles to carry these vibrations, so sound cannot propagate.
A common demonstration: place a ringing bell inside a glass jar connected to a vacuum pump. Initially, listeners outside the jar hear the bell clearly because air inside transmits sound. As the pump removes air, the sound gradually becomes fainter and finally almost disappears when most air is removed.
Observation: the bell still vibrates inside the jar, but sound is not heard outside because there is no medium to transmit the vibrations to the listener’s ear.
Conclusion: this shows that the presence of particles in a medium is essential for sound to travel. It also distinguishes vibration of an object from perception of sound, which requires a transmitting medium.
Q3. Describe the relationship between frequency and pitch. Use examples such as bird chirping, tuning forks, and stringed instruments to explain how frequency determines pitch in everyday sounds.
Answer:
Frequency is the number of vibrations per second measured in Hertz (Hz). Pitch is how high or low a sound seems to our ear. The general rule is: higher frequency → higher pitch, and lower frequency → lower pitch.
Example: small birds chirp at high frequencies, so their sounds have a high pitch. A tuning fork of 512 Hz will produce a specific, clear pitch; another fork with a lower Hz value produces a lower pitch.
In stringed instruments, changing the length, tension, or mass per unit length of the string changes its frequency. Shortening a string or increasing its tension raises the frequency and therefore raises the pitch.
Our ears interpret the frequency pattern of incoming sound waves as pitch; thus musical notes are distinguished mostly by their frequency values.
Q4. Explain how amplitude affects loudness. Include real-life examples (whisper vs. shout, concerts) and mention one way to measure loudness scientifically.
Answer:
Amplitude refers to the maximum displacement of particles from their rest position in a sound wave. Loudness is how strong or soft a sound appears to our ear and is related to amplitude: greater amplitude → louder sound; smaller amplitude → softer sound.
Examples: a whisper has very small amplitude and is heard only nearby. A shout or a struck drum has large amplitude and can be heard far away. At concerts, amplifiers increase the amplitude of the sound waves so the music reaches the audience loudly.
Scientifically, loudness is measured in decibels (dB), which represents the sound intensity level. A difference of about 10 dB is usually perceived by humans as roughly twice or half as loud.
Thus, amplitude determines the energy carried by the wave, and higher energy produces greater perceived loudness.
Q5. Explain why sound travels at different speeds in solids, liquids, and gases. Give examples to support your explanation, such as hearing a train through rails and hearing sounds underwater.
Answer:
Sound travels by particle-to-particle interaction. The speed of sound depends on how close the particles are and how strongly they interact (elasticity). In solids, particles are tightly packed and bonded strongly, so vibrations pass quickly; hence sound speed is highest in solids. In liquids, particles are less tightly packed than solids but closer than gases, so speed is intermediate. In gases, particles are far apart and collisions are less frequent, so sound travels slowest.
Example: when you place your ear on a railway track you can hear a distant train earlier and more clearly because sound travels faster through the metal (solid) than through air.
Underwater, sound travels faster and farther than in air because water is denser and transmits vibrations more efficiently; that is why you can hear a boat engine well while submerged.
Therefore, particle spacing and elasticity of the medium determine the relative speeds of sound.
High Complexity (Analytical & Scenario-Based)
Q6. You strike a tuning fork and touch it to a wooden table; you also strike another tuning fork and immerse it in water. Explain in detail what happens in each case and why the loudness differs between the two situations.
Answer:
When the tuning fork is struck, its prongs vibrate at a fixed frequency. If you touch it to a wooden table, vibrations transfer from the fork to the table. The table has a much larger surface area and moves more air, acting as a soundboard, so the sound becomes louder and richer. This is because energy spreads into the table and then into the air more efficiently than from the small fork alone.
When you immerse a struck tuning fork in water, vibrations couple into the water. Water is a good medium for transmitting sound, so some sound goes into the water. However, if your ear is in air and the fork is partly submerged, much of the sound energy remains in water and does not transfer back efficiently into air, so the sound heard in air may be weaker. Conversely, an underwater listener would hear it louder.
Thus, matching of mediums and surface area for transmitting vibrations matters: a solid table amplifies sound in air, while water transmits sound well but may not radiate it into air efficiently.
Q7. Design and describe a classroom experiment (materials, method, observations, and conclusion) to show that sound cannot travel in vacuum. Include expected observations at each step.
Answer:
Materials: a small electric bell or mechanical bell, a glass bell jar with a removable lid, vacuum pump, connecting tube, and power source for the bell.
Method: place the ringing bell under the bell jar and switch it on so it rings continuously. With the jar filled with air, students outside can hear the bell clearly. Start the vacuum pump to slowly remove air from the jar while the bell keeps ringing. Ask students to observe the sound intensity at intervals as pressure decreases. Continue until a near vacuum is reached.
Observations: initially the bell is loud. As air is pumped out, the sound becomes fainter. Eventually, when most air is removed, the bell may still be seen vibrating but is almost inaudible outside the jar. If air is reintroduced, the sound becomes loud again.
Conclusion: Sound intensity depends on the presence of a medium. In near vacuum, sound cannot travel to the listener, showing that a medium is necessary for sound propagation. This demonstration separates the physical vibration of the source from the propagation of sound waves.
Q8. A guitar string produces a note of a certain pitch. Explain analytically how changing the length, tension, and mass per unit length of the string affects its frequency and therefore its pitch. Suggest practical ways a guitarist changes pitch.
Answer:
The frequency of a stretched string depends on its length (L), tension (T), and mass per unit length (μ). In simple terms: shorter length → higher frequency; greater tension → higher frequency; larger mass per unit length → lower frequency.
Why that happens: a shorter string vibrates faster because each wave cycle takes less time to travel along the string. Increasing tension makes the string tighter, so waves travel faster along it and frequency increases. Increasing the string’s mass per unit length makes it heavier and harder to accelerate, so it vibrates more slowly and frequency decreases.
Practical ways a guitarist changes pitch: pressing the string against a fret shortens its vibrating length; tuning pegs increase or decrease tension; using a thicker or thinner string changes mass per unit length. These controls let the player produce different musical notes by adjusting the string’s physical properties.
Q9. Define an echo and explain the conditions necessary to hear a clear echo. If the speed of sound in air is 340 m/s and the minimum time gap required for a person to distinguish an echo is 0.1 s, calculate the minimum distance of the reflecting surface from the source. Also explain how reverberation differs from echo.
Answer:
An echo is the reflection of sound from a surface that returns after a delay, allowing the listener to hear the original sound and the reflection separately. To hear a clear echo, there must be a large, hard, and smooth reflecting surface located sufficiently far from the source so that the reflected sound arrives after a ...
