1.0 Production and Propagation of Sound
Sound is a form of energy that produces the sensation of hearing in our ears. It is produced by vibrating bodies. When a body vibrates, it transfers its energy to the surrounding medium, which then travels in the form of a wave.
Nature of Sound Waves
- Mechanical Wave: Sound requires a material medium (solid, liquid, or gas) to travel. It cannot travel through a vacuum.
- Longitudinal Wave: Particles of the medium vibrate back and forth parallel to the direction of propagation of the wave.
1.1 Mechanism of Propagation
As a sound wave moves through air, it creates regions of high and low pressure:
- Compressions: Regions where particles are crowded together (High Pressure).
- Rarefactions: Regions where particles are spread apart (Low Pressure).
1.2 Speed of Sound
The speed of sound depends on the elasticity and density of the medium. Generally:
$V_{solids} > V_{liquids} > V_{gases}$
- In Air: Approx. $330\,m/s$ to $340\,m/s$ (at room temp).
- In Water: Approx. $1500\,m/s$.
- In Steel: Approx. $5000\,m/s$.
The Wave Equation
$$V = f \times \lambda$$
Where: $V$ = Velocity, $f$ = Frequency (Hz), $\lambda$ = Wavelength (m).
This famous experiment proves that sound cannot travel in a vacuum. As air is pumped out of a jar containing a ringing electric bell, the sound becomes fainter and eventually disappears, even though the hammer is still seen striking the bell.
A tuning fork has a frequency of $256\,Hz$. If the speed of sound in air is $340\,m/s$, calculate the wavelength of the sound produced.
Solution:
1. Given: Frequency ($f$) = $256\,Hz$, Velocity ($V$) = $340\,m/s$.
2. Formula: $\lambda = V / f$.
3. Calculation: $\lambda = 340 / 256 = \mathbf{1.328\,m}$.
Final Answer: The wavelength is approximately $1.33\,m$.
Lightning is seen much earlier than thunder is heard because the speed of light ($3 \times 10^8\,m/s$) is nearly a million times faster than the speed of sound ($340\,m/s$)!
2.0 Characteristics of Sound
Every sound we hear—from a whisper to a jet engine—can be distinguished using three fundamental characteristics. These allow our brain to process the difference between a guitar and a piano, or a high-pitched whistle and a low-pitched drum.
1. Loudness (Intensity)
Loudness is the sensation produced in the ear which enables us to distinguish between a faint sound and a loud sound. It depends on the Amplitude of the vibration.
- Unit: Decibel (dB).
- Factor: $\text{Loudness} \propto (\text{Amplitude})^2$.
2. Pitch (Frequency)
Pitch is that characteristic of sound by which we can distinguish between a shrill sound and a grave (flat) sound. It depends on the Frequency of vibration.
- High frequency = High pitch (e.g., voice of a child or a woman).
- Low frequency = Low pitch (e.g., voice of a man or a bass drum).
3. Quality or Timbre
Quality is the characteristic that allows us to distinguish between sounds of the same pitch and loudness produced by two different instruments. It depends on the Waveform (the presence of overtones or harmonics).
Infrasonic and Ultrasonic Sounds
The Human Audible Range: $20\,Hz$ to $20,000\,Hz$
Infrasonic: Frequency $< 20\,Hz$ (e.g., Earthquakes).
Ultrasonic: Frequency $> 20,000\,Hz$ (e.g., Bats, Ultrasound scans).
Intensity is an objective physical quantity (Energy per unit area per unit time). Loudness is a subjective sensation that varies from person to person depending on the sensitivity of their ears.
If the amplitude of a sound wave is tripled, by what factor does its loudness increase?
Solution:
1. We know that $\text{Loudness} \propto (\text{Amplitude})^2$.
2. New Amplitude = $3 \times \text{Original Amplitude}$.
3. New Loudness $\propto (3)^2 = 9$.
Final Answer: The loudness increases by a factor of 9.
Bats don't "see" with their eyes in the dark; they use Echolocation! They emit ultrasonic squeaks that bounce off insects and walls. By hearing the reflection, they can map out their surroundings with incredible precision.
3.0 Reflection of Sound and Echoes
Just like light, sound waves also bounce back when they strike a hard surface. This return of sound into the same medium is called the Reflection of Sound. It follows the same laws as light: the angle of incidence equals the angle of reflection.
What is an Echo?
An Echo is the sound heard after reflection from a distant obstacle (such as a cliff or a tall building) after the original sound has ceased.
Condition for Echo: The sensation of sound persists in our brain for about 0.1 seconds (Persistence of Hearing). To hear a distinct echo, the reflected sound must reach the ear after this interval.
The Echo Formula
Total distance traveled by sound is $2d$ (to the obstacle and back):
$$V = \frac{2d}{t} \quad \Rightarrow \quad d = \frac{V \times t}{2}$$
Where $V$ = Velocity of sound, $d$ = distance to obstacle, $t$ = time taken.
3.1 Minimum Distance for Echo
If we take the speed of sound in air as $340\,m/s$ and the minimum time for persistence of hearing as $0.1\,s$:
$d = \frac{340 \times 0.1}{2} = \mathbf{17\,meters}$
Therefore, to hear a distinct echo in air, the minimum distance between the source of sound and the reflecting surface must be 17 m.
3.2 Reverberation
In a small room or a hall, the reflecting surfaces are closer than 17 m. The reflected sounds reach the ear in less than 0.1s and mix with the original sound. This results in a prolonged, blurred sound called Reverberation. To reduce this in auditoriums, walls are covered with sound-absorbing materials like thermocol or heavy curtains.
Since the speed of sound in water is much higher ($\approx 1500\,m/s$), the minimum distance required to hear an echo underwater is much greater:
$d = \frac{1500 \times 0.1}{2} = \mathbf{75\,meters}$.
A person stands between two parallel cliffs and fires a gun. He hears the first echo after 2s and the second echo after 3s. If the speed of sound is $340\,m/s$, find the distance between the cliffs.
Solution:
1. Distance to Cliff 1 ($d_1$): $\frac{V \times t_1}{2} = \frac{340 \times 2}{2} = 340\,m$.
2. Distance to Cliff 2 ($d_2$): $\frac{V \times t_2}{2} = \frac{340 \times 3}{2} = 510\,m$.
3. Total Distance: $d_1 + d_2 = 340 + 510 = \mathbf{850\,meters}$.
Final Answer: The cliffs are $850\,m$ apart.
The Gol Gumbaz in Bijapur, India, is famous for its "Whispering Gallery." Because of its unique hemispherical shape, a sound produced at one end is reflected multiple times, allowing even a faint whisper to be heard clearly across the massive dome!
4.0 Applications of Ultrasound: SONAR
SONAR stands for Sound Navigation and Ranging. It is a technique that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects such as shipwrecks, shoals of fish, or the ocean floor.
How SONAR Works
- A Transmitter sends out ultrasonic pulses into the water.
- These waves travel through the water, strike an object, and get reflected back.
- A Detector (Receiver) picks up the reflected waves and converts them into electrical signals.
- The time interval ($t$) between transmission and reception is recorded.
Echo-Ranging Calculation
The depth of the sea ($d$) is calculated using:
$$d = \frac{V \times t}{2}$$
Ultrasound is used because it has a high frequency, small wavelength, and can travel long distances in water without spreading out.
4.1 Noise Pollution
While music is pleasing to the ear, Noise is defined as any unpleasant, loud, or jarring sound that is produced by irregular vibrations. Continuous exposure to high noise levels leads to Noise Pollution.
| Aspect | Details |
|---|---|
| Threshold of Pain | Sounds above 120 dB become physically painful. |
| Harmful Effects | Hearing loss, hypertension (high BP), stress, and lack of sleep (insomnia). |
| Control Measures | Using silencers in engines, planting trees (which act as sound buffers), and restricted use of loudspeakers. |
Just like SONAR, doctors use ultrasonic waves to get images of the heart. This is called Echocardiography. Since ultrasound is non-invasive and safer than X-rays, it is also used to monitor the growth of a fetus during pregnancy.
A SONAR device on a submarine sends out a signal and receives an echo 5 seconds later. If the speed of sound in water is $1500\,m/s$, calculate the distance of the object from the submarine.
Solution:
1. Total time ($t$): $5\,s$.
2. Speed ($V$): $1500\,m/s$.
3. Formula: $d = (V \times t) / 2$.
4. Calculation: $d = (1500 \times 5) / 2 = 7500 / 2 = \mathbf{3750\,m}$.
Final Answer: The object is at a distance of $3750\,m$ (or $3.75\,km$).
A "Sonic Boom" occurs when an airplane flies faster than the speed of sound! It creates shockwaves in the air that sound like a massive explosion and can even shatter window panes on the ground.