Sound

🧠 Answer:
The vibrating object causes particles of the medium (like air) to vibrate. These vibrations travel in the form of compressions and rarefactions, creating a longitudinal sound wave. The wave reaches our ears, causing our eardrum to vibrate, which is then processed by our brain as sound.

🧠 Answer:
When the school bell is struck, it begins to vibrate. These vibrations disturb the surrounding air particles, creating compressions and rarefactions. These sound waves travel through the air to our ears, and we hear the bell ringing.

🧠 Answer:
Sound waves are called mechanical waves because they require a material medium (like air, water, or solids) to travel. They cannot travel in a vacuum, which is a property of mechanical waves.

🧠 Answer:
No, you will not be able to hear your friend on the moon. This is because the moon has no atmosphere, so there is no medium for sound to travel. Sound cannot travel in a vacuum.

🧠 Answer:
(a) Loudness depends on the amplitude of the wave. Greater amplitude means louder sound.
(b) Pitch depends on the frequency of the wave. Higher frequency means higher pitch.

🧠 Answer:
A guitar has a higher pitch than a car horn. Guitar strings vibrate at a higher frequency, producing a sharper and higher-pitched sound.

🧠 Answer:
– Wavelength (λ): The distance between two consecutive compressions or rarefactions.
– Frequency (f): Number of vibrations per second (unit: Hertz, Hz).
– Time Period (T): Time taken for one complete vibration (T = 1/f).
– Amplitude: Maximum displacement of particles from their mean position. Greater amplitude means louder sound.

🧠 Answer:
The speed of sound is the product of its frequency and wavelength.
Formula:

Speed (v) = Frequency (f) × Wavelength (λ)

📘 Given:
Frequency (f) = 220 Hz
Speed (v) = 440 m/s

📘 Formula:

Wavelength (λ) = v / f
              = 440 / 220
              = 2 m

✅ Final Answer: Wavelength = 2 metres

🧠 Answer:
The time interval between successive compressions is the time period (T) of the wave.

📘 Given: Frequency (f) = 500 Hz

📘 Formula:

Time period (T) = 1 / f
              = 1 / 500
              = 0.002 seconds

✅ Final Answer: 0.002 seconds

🧠 Answer:

🧠 Answer:
Sound travels the fastest in iron, then in water, and the slowest in air at a given temperature.

✅ Order of speed:
Iron > Water > Air

📊 Typical Speeds:

• In air (at 25°C): ~346 m/s
• In water: ~1500 m/s
• In iron: ~5000 m/s

📘 Given:
Time (t) = 3 seconds
Speed of sound (v) = 342 m/s

🧠 Explanation:
In echo problems, the sound travels to the surface and back. So the time includes the round trip.

📘 Formula:

Distance = (v × t) / 2
         = (342 × 3) / 2
         = 1026 / 2
         = 513 m

✅ Final Answer: Distance of reflecting surface = 513 metres

🧠 Answer:
The ceilings of concert halls are curved so that sound reflects uniformly in all directions. This ensures that sound reaches every corner of the hall clearly. Curved surfaces help in the even distribution of sound by reflecting it toward the audience, improving audibility and clarity.

🧠 Answer:
The audible range of the average human ear is from 20 Hz to 20,000 Hz (20 kHz).

🧠 Answer:

• (a) Infrasound: Frequencies below 20 Hz
• (b) Ultrasound: Frequencies above 20,000 Hz (20 kHz)

🧠 Answer:
Sound is a form of energy produced by vibrating objects. When an object vibrates, it disturbs the particles of the surrounding medium (like air), creating sound waves that travel to our ears and make us hear the sound.

🧠 Answer:
When a vibrating object moves forward, it pushes air particles together creating a compression (region of high pressure). When it moves backward, it leaves a space between particles, creating a rarefaction (region of low pressure). These alternate compressions and rarefactions travel in the form of sound waves.

Diagram: Use the 16:9 labeled diagram showing sound propagation with compressions and rarefactions (as created earlier).

🧠 Answer:
In a sound wave, the particles of the medium vibrate parallel to the direction of wave propagation. Since this is the defining property of longitudinal waves, sound is called a longitudinal wave.

🧠 Answer:
The quality or timbre of sound helps us identify different voices. Even if pitch and loudness are the same, each person’s voice has a unique quality which allows us to recognize it.

🧠 Answer:
Light travels much faster than sound. So, even though thunder and lightning occur at the same time, the light reaches us first, and the sound reaches a few seconds later, causing the delay in hearing the thunder.

📘 Given: Speed of sound = 344 m/s Frequency range = 20 Hz to 20,000 Hz (20 kHz)

Formula:

Wavelength = Speed / Frequency

🧮 Calculation:

For 20 Hz:
λ = 344 / 20 = 17.2 m

For 20,000 Hz:
λ = 344 / 20000 = 0.0172 m

✅ Final Answer:
Wavelength range = 0.0172 m to 17.2 m

📘 Given:
Speed of sound in aluminium ≈ 6420 m/s
Speed of sound in air ≈ 343 m/s

🧠 Concept:
Time is inversely proportional to speed (t ∝ 1/v)

Ratio = time in air / time in aluminium
      = speed in aluminium / speed in air
      = 6420 / 343 ≈ 18.72

✅ Final Answer: Ratio = 18.72 : 1

📘 Given:
Frequency = 100 Hz = 100 vibrations per second
Time = 1 minute = 60 seconds

Number of vibrations = 100 × 60 = 6000

✅ Final Answer: 6000 vibrations in a minute

🧠 Answer:
Yes, sound follows the same laws of reflection as light:

• Angle of incidence = Angle of reflection
• Incident wave, reflected wave, and normal lie in the same plane

🧠 Answer:
Yes, on a hotter day the speed of sound increases because air becomes less dense. So the sound takes less time to return, and an echo may be heard more distinctly if the time gap is ≥ 0.1 s.

🧠 Answer:

• Used in SONAR to detect underwater objects.
• Used in stethoscopes by doctors to hear heartbeats.
• Used in the design of auditoriums and concert halls for better acoustics.

📘 Given:
Height (h) = 500 m
g = 10 m/s²
Speed of sound (v) = 340 m/s

Step 1: Time taken to fall

t₁ = √(2h/g) = √(2×500 / 10) = √100 = 10 s

Step 2: Time for sound to travel back

t₂ = Distance / Speed = 500 / 340 ≈ 1.47 s

✅ Final Answer:
Splash is heard after 10 + 1.47 = 11.47 seconds

📘 Given:
Speed (v) = 339 m/s
Wavelength (λ) = 1.5 cm = 0.015 m

📘 Formula:

Frequency (f) = v / λ
             = 339 / 0.015
             ≈ 22600 Hz

✅ Final Answer:
Frequency ≈ 22,600 Hz → This is not audible as it is above 20,000 Hz (ultrasound).

🧠 Answer:
Reverberation is the prolonged persistence of sound due to multiple reflections from surfaces like walls, ceiling, and floor. It makes sound unclear or noisy in a hall.

To reduce reverberation:

• Use sound-absorbing materials (curtains, carpets)
• Install false ceilings with acoustic tiles
• Add padded furniture to absorb sound

🧠 Answer:
Loudness is the perception of sound intensity by the human ear. It tells how strong or weak a sound is.

It depends on:

• Amplitude of the sound wave (more amplitude → more loudness)
• Sensitivity of the listener’s ear

✅ Unit of Loudness: Decibel (dB)

🧠 Answer:
Ultrasound waves are used to clean delicate objects like watches, electronic components, or surgical instruments. The high-frequency waves produce vibrations in a cleaning fluid, loosening dirt particles even from tiny holes or corners.

🧠 Answer:
Ultrasound waves are passed through a metal block. If there is a defect or crack, part of the wave is reflected back early. The remaining wave passes through. By analyzing the reflected waves, technicians can locate hidden flaws without cutting the metal.