1.0 Production of Sound
Sound is a form of energy that produces the sensation of hearing in our ears. In Physics, we learn that sound is always produced by a vibrating body. When an object moves back and forth rapidly, it disturbs the surrounding air particles, creating sound waves.
What are Vibrations?
A vibration is a rapid "to-and-fro" or "back-and-forth" motion of an object about its mean position. Without vibration, there is no sound.
- Stretched Strings: Guitar, Violin.
- Stretched Membranes: Drums, Tabla.
- Air Columns: Flute, Whistle.
1.1 Propagation of Sound
Unlike light, sound cannot travel through a vacuum. It requires a material medium (solid, liquid, or gas) to travel from one point to another. This is because sound travels as a mechanical wave, passing energy through the collision of particles.
Speed of Sound Comparison
Solids > Liquids > Gases
Sound travels fastest in solids because particles are closely packed.
1.2 Terms Related to Sound
To describe sound waves mathematically, we use several key terms:
- Amplitude ($A$): The maximum displacement of a vibrating particle from its mean position. It determines the Loudness.
- Frequency ($f$): The number of vibrations per second. It determines the Pitch.
Unit: Hertz (Hz) - Time Period ($T$): The time taken to complete one full vibration.
If an electric bell is placed inside a jar and the air is gradually pumped out (creating a vacuum), the sound of the bell becomes fainter and eventually disappears, even though the hammer is still striking. This proves sound needs a medium.
A tuning fork vibrates 500 times in 2 seconds. Calculate its frequency and time period.
Solution:
1. Frequency ($f$): Number of vibrations / Time = $500 / 2 = 250\,Hz$.
2. Time Period ($T$): $1 / f = 1 / 250 = 0.004\,s$.
Final Answer: Frequency is $250\,Hz$ and Time Period is $0.004\,seconds$.
Lightning and thunder happen at the same time, but we see the flash first and hear the sound later. This is because light travels at 300,000 km/s, while sound travels at only about 0.34 km/s!
2.0 Characteristics of Sound
Even if two sounds have the same loudness, we can distinguish between a guitar and a flute, or between a man's voice and a woman's voice. This is because sound has three primary characteristics: Loudness, Pitch, and Quality.
2.1 Loudness
Loudness is the characteristic that distinguishes a faint sound from a loud one. It depends primarily on the Amplitude of the vibration. The larger the amplitude, the louder the sound.
- Unit: Decibel (dB).
- Factors: Distance from the source, surface area of the vibrating body, and density of the medium.
2.2 Pitch (Shrillness)
Pitch is the characteristic that distinguishes a shrill sound from a grave (flat) one. It depends on the Frequency of the vibration. A woman's voice generally has a higher frequency and thus a higher pitch than a man's voice.
2.3 Quality (Timbre)
Quality is the characteristic that allows us to distinguish between sounds produced by different instruments (like a piano vs. a violin) even if they have the same loudness and pitch. This is due to the different "waveforms" produced by different sources.
The Core Relation
Loudness $\propto (Amplitude)^2$
Pitch $\propto Frequency$
3.0 Range of Hearing
Human ears are not designed to hear all frequencies of sound. We can only hear sound within a specific range called the Audible Range.
- Infrasonic Sound: Frequencies below 20 Hz (e.g., produced by earthquakes, whales).
- Audible Range: 20 Hz to 20,000 Hz (20 kHz).
- Ultrasonic Sound: Frequencies above 20,000 Hz (e.g., used by bats, ultrasound machines).
Unpleasant and loud sounds are called Noise. Continuous exposure to sounds above 80 dB can lead to permanent hearing loss, stress, and high blood pressure.
If the amplitude of a vibrating object is tripled, how will its loudness change?
Solution:
1. Rule: Loudness is proportional to the square of the amplitude.
2. Calculation: $(3)^2 = 9$.
Final Answer: The loudness will become 9 times the original loudness.
SONAR (Sound Navigation and Ranging) uses ultrasonic waves to measure the depth of the sea and to locate underwater objects like submarines or shipwrecks.
4.0 Reflection of Sound
Just like light, sound waves also bounce back when they strike a hard surface. This phenomenon is called the Reflection of Sound. It follows the same laws as light reflection: the angle of incidence equals the angle of reflection.
4.1 Echo
An Echo is the repetition of sound heard after it is reflected from a distant hard surface (like a cliff, a wall, or a mountain).
Conditions for Hearing an Echo
- Persistence of Hearing: Our brain retains a sound for about 0.1 seconds. The reflected sound must reach our ears after 0.1s to be heard as a distinct echo.
- Minimum Distance: Since the speed of sound in air is approx. $340\,m/s$, the sound must travel $34\,m$ ($340 \times 0.1$) to and fro. Thus, the minimum distance of the reflector must be 17 metres.
- Intensity: The sound must be loud enough to be heard after reflection.
Echo Calculation Formula
$$2d = v \times t$$
Where: $d$ = distance to reflector, $v$ = speed of sound, $t$ = time taken for echo.
4.2 Absorption of Sound
Not all surfaces reflect sound. Soft, porous, or uneven surfaces like carpets, curtains, and thermocol tend to absorb sound energy. This is useful in cinema halls and auditoriums to prevent Reverberation (prolonged echoing).
An Echo is a single distinct reflection. Reverberation is the result of multiple reflections in a closed space that make the sound "linger" or become blurred. We use sound absorbers to reduce reverberation.
A man fires a gun in front of a building and hears the echo after 2 seconds. If the speed of sound is 340 m/s, calculate his distance from the building.
Solution:
1. Given Time ($t$): $2\,s$
2. Speed of sound ($v$): $340\,m/s$
3. Formula: $2d = v \times t$
4. Calculation: $2d = 340 \times 2 = 680\,m$
5. Distance ($d$): $680 / 2 = 340\,m$
Final Answer: The man is $340\,m$ away from the building.
A Stethoscope works on the principle of multiple reflections of sound. The heartbeats of the patient travel through the tube to the doctor's ears by bouncing off the inner walls of the tube.