1.0 Reflection of Sound and Echoes
Just like light, sound waves also obey the laws of reflection when they strike a surface. However, because sound waves have much larger wavelengths than light, they do not require a polished surface to reflect; even a rough wall or a cliff can act as a reflector.
What is an Echo?
An Echo is the sound heard after reflection from a distant obstacle (such as a cliff, a wall, or a mountainside) after the original sound has ceased.
Conditions for hearing an Echo:
- Persistence of Hearing: The sensation of sound persists in our brain for about 0.1 seconds. The reflected sound must reach the ear after this interval.
- Minimum Distance: Taking the speed of sound as $340\,m/s$, the minimum distance between the source and the obstacle must be 17 metres.
- Intensity: The sound must be loud enough to be heard after reflection.
The Echo Formula
If $d$ is the distance to the obstacle, the sound travels $2d$ (to the obstacle and back):
$$V = \frac{2d}{t} \quad \text{or} \quad d = \frac{V \times t}{2}$$
Where: $V$ = speed of sound, $t$ = time taken to hear the echo.
1.1 Reverberation vs. Echo
It is important to distinguish between a distinct echo and the "rolling" sound heard in large halls:
- Echo: A distinct, separate sound heard after the original sound.
- Reverberation: If the reflecting surface is closer than 17m, the reflected sound merges with the original sound, causing it to be prolonged. This is common in auditoriums and is controlled using sound-absorbent materials.
The minimum distance of 17m is calculated for dry air at room temperature. Since the speed of sound increases with temperature and humidity, the minimum distance required to hear a clear echo will also increase on a hot or humid day.
A person fires a gun in front of a cliff and hears an echo after $2.5$ seconds. If the speed of sound is $340\,m/s$, calculate the distance of the cliff from the person.
Solution:
1. Given: $V = 340\,m/s$, $t = 2.5\,s$.
2. Formula: $d = (V \times t) / 2$.
3. Calculation: $d = (340 \times 2.5) / 2 = 850 / 2 = \mathbf{425\,m}$.
Final Answer: The cliff is at a distance of $425\,m$.
Bats and Dolphins use Echolocation to navigate and hunt! They emit ultrasonic waves and listen for the echoes. By calculating the time delay, their brains can "see" the distance and size of obstacles or prey even in complete darkness.
2.0 Applications of Echoes: SONAR and Ultrasound
The principle of echoes is not just a curiosity of nature; it is a powerful tool used in navigation, defense, and medicine. By using Ultrasonic waves (sound waves with frequency > 20,000 Hz), we can "see" objects that are invisible to the naked eye.
SONAR (Sound Navigation and Ranging)
SONAR is a technique used in ships and submarines to detect and locate underwater objects like shoals of fish, shipwrecks, or enemy submarines.
- Mechanism: A transmitter sends an ultrasonic pulse into the sea. The pulse reflects off the object and is received by a detector.
- Why Ultrasound? Ultrasonic waves can travel long distances in water without being scattered and are highly directional.
2.1 Medical Applications of Ultrasound
In the field of medicine, the use of ultrasound for diagnostic purposes is known as Ultrasonography. It is preferred over X-rays for many internal examinations because it is non-invasive and does not involve harmful radiation.
- Echocardiography: Used to create images of the heart and its valves to detect abnormalities.
- Ultrasonography: Used to monitor the growth of a fetus during pregnancy or to detect stones in the gallbladder or kidneys.
- Lithotripsy: High-intensity ultrasonic waves are used to break kidney stones into fine grains that can be passed out through urine.
Depth of Sea Formula
If the speed of ultrasound in water is $V$ and the time interval between transmission and reception is $t$:
$$\text{Depth } (d) = \frac{V \times t}{2}$$
Infrasound (frequency < 20 Hz) has very long wavelengths. These waves diffract (bend) around obstacles rather than reflecting off them. To get a sharp echo and precise location, we need Ultrasound because its short wavelength allows it to reflect clearly even from small objects.
An ultrasonic wave is sent from a ship to the bottom of the sea. It is received back after $4$ seconds. If the speed of sound in seawater is $1500\,m/s$, what is the depth of the sea?
Solution:
1. Given: $V = 1500\,m/s$, $t = 4\,s$.
2. Formula: $d = (V \times t) / 2$.
3. Calculation: $d = (1500 \times 4) / 2 = 6000 / 2 = \mathbf{3000\,m}$.
Final Answer: The depth of the sea is $3000\,m$ (or $3\,km$).
Ultrasound is also used in industries to clean parts located in hard-to-reach places, such as spiral tubes or electronic components. The objects are placed in a cleaning solution, and ultrasonic waves create high-pressure bubbles that "scrub" the surfaces clean!
3.0 Natural, Damped, and Forced Vibrations
Objects in our world are constantly vibrating. However, the way they vibrate depends on whether they are left alone, subjected to friction, or pushed by an external force. In ICSE Physics, understanding these distinctions is key to explaining how musical instruments and machines work.
1. Natural (Free) Vibrations
Vibrations of a body in the absence of any external force or resistance. The body vibrates with its own natural frequency.
- Amplitude: Remains constant over time.
- Environment: Possible only in a vacuum.
2. Damped Vibrations
Vibrations where the amplitude decreases gradually with time due to external resistive forces (like air friction).
- Energy: Lost to the surroundings as heat.
- Frequency: Slightly less than the natural frequency.
3.1 Forced Vibrations
When a body is made to vibrate by an external periodic force, its vibrations are called Forced Vibrations. In this state, the body does not vibrate with its natural frequency; it is forced to vibrate with the frequency of the external force.
| Feature | Natural Vibrations | Forced Vibrations |
|---|---|---|
| Frequency | Natural frequency of the body. | Frequency of the external force. |
| Amplitude | Depends on initial push. | Depends on the difference in frequencies. |
Natural Frequency of a Stretched String
For a string of length $l$, tension $T$, and mass per unit length $m$:
$$f = \frac{1}{2l} \sqrt{\frac{T}{m}}$$
A special case of forced vibrations occurs when the frequency of the external force becomes equal to the natural frequency of the body. This is called Resonance, resulting in vibrations with a very large amplitude. We will cover this in detail in the next section!
Why does the sound from a tuning fork become louder when its stem is pressed against a tabletop?
Solution:
1. When the tuning fork is held in the air, only a small amount of air is set into vibration.
2. When pressed on the table, the tuning fork forces the large surface area of the tabletop to vibrate at the same frequency.
3. A large volume of air is set into forced vibration, resulting in a much louder sound.
Final Answer: Due to the forced vibrations of the larger surface area of the table.
The suspension system (shock absorbers) in a car is designed to dampen vibrations. Without them, hitting a single bump would make your car bounce up and down with "natural vibrations" for a very long time, making for a very uncomfortable ride!
4.0 Resonance
Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body begins to vibrate with a very large amplitude. This phenomenon is a frequent topic for reasoning-based questions in the ICSE Physics paper.
Condition for Resonance
Resonance occurs only when:
Applied Frequency ($f$) = Natural Frequency ($f_0$)
- Result: A loud sound is produced because the amplitude of vibration becomes maximum.
4.1 Famous Examples of Resonance
You may encounter these examples in the "Give Reason" section of your exam:
- Troops Crossing a Bridge: Soldiers are ordered to break their step (not march in unison) while crossing a bridge. If the frequency of their rhythmic marching coincides with the natural frequency of the bridge, resonance may occur, causing the bridge to vibrate with large amplitude and potentially collapse.
- Radio/TV Tuning: When you turn the knob of a radio, you are changing the natural frequency of its electrical circuit. When this matches the frequency of the incoming signal from a station, electrical resonance occurs, and you hear that specific station clearly.
- Rattle in a Car: Sometimes, a specific part of a car (like a loose window) vibrates loudly only at a certain speed. This happens because the frequency of the engine's vibrations matches the natural frequency of that part.
Resonance in Air Columns
When a vibrating tuning fork is held over a resonance tube, a loud sound is heard at a specific water level ($l$):
$$V = 4f(l + e)$$
Where: $V$ = Speed of sound, $f$ = frequency, $e$ = end correction.
Remember: All resonance is a forced vibration, but not all forced vibrations are resonance. In forced vibration, the body can vibrate at any frequency imposed on it. In resonance, it must match the natural frequency to achieve large amplitude.
Two tuning forks A and B of frequencies $256\,Hz$ and $512\,Hz$ respectively are vibrated. Which one will produce resonance in a sonometer wire of natural frequency $256\,Hz$?
Solution:
1. Natural frequency of wire = $256\,Hz$.
2. Frequency of Fork A = $256\,Hz$; Frequency of Fork B = $512\,Hz$.
3. Since the frequency of Fork A matches the natural frequency of the wire, Fork A will cause resonance.
Final Answer: Tuning fork A.
An opera singer can shatter a wine glass by singing a very high note! If the singer hits a pitch that exactly matches the natural frequency of the glass, the resulting resonance makes the glass vibrate so violently that it breaks apart.
5.0 Characteristics of Sound
Two sounds can be distinguished from each other based on three main characteristics: Loudness, Pitch, and Quality (Timbre). Even if two musical instruments play the same note, our ears can tell them apart because of these factors.
5.1 Loudness and Intensity
Loudness is the sensation felt by the ear, which depends on the energy of the sound wave. It is primarily determined by the Amplitude of the vibration.
- Unit: The S.I. unit of intensity is $Watt/m^2$, but loudness is expressed in decibels (dB).
- Factors: Loudness $\propto (\text{Amplitude})^2$. It also depends on the surface area of the vibrating body and the distance from the source.
5.2 Pitch and Frequency
Pitch is the characteristic of sound that distinguishes a "shrill" sound from a "grave" (flat) sound. It is determined by the Frequency of the vibration.
- A lady's voice or a humming bee has a High Pitch (High Frequency).
- A man's voice or a roaring lion has a Low Pitch (Low Frequency).
5.3 Quality or Timbre
This characteristic allows us to distinguish between sounds of the same pitch and loudness produced by different instruments (like a piano and a violin). It depends on the Waveform (the presence of subsidiary vibrations called overtones).
| Characteristic | Subjective Nature | Objective Measurable Quantity |
|---|---|---|
| Loudness | Loud or Soft | Amplitude / Intensity |
| Pitch | Shrill or Grave | Frequency |
| Quality | Musical Tone | Waveform |
This is a common "Distinguish between" question. Intensity is an objective physical quantity (energy per unit area per unit time), while 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 doubled, by what factor does the loudness increase?
Solution:
1. We know that Intensity (and hence Loudness) is directly proportional to the square of the amplitude ($I \propto A^2$).
2. If the new amplitude $A' = 2A$.
3. New Loudness $\propto (2A)^2 = 4A^2$.
Final Answer: The loudness increases by 4 times.
Sound over 120 dB can cause physical pain to the human ear and permanent hearing damage if exposure is prolonged. A jet engine at take-off is about 140 dB—this is why ground crew always wear heavy-duty ear protection!