Sound

Sound:
Sound is a form of energy that produces the sensation of hearing in our ears. It is a mechanical wave that requires a material medium (solid, liquid, or gas) for its propagation.

How sound is produced:
Sound is produced by the vibration of objects. When an object vibrates, it causes the particles of the surrounding medium to vibrate. These vibrations travel through the medium in the form of a wave, which we perceive as sound.

Example: When we pluck a stretched rubber band, it vibrates and produces sound. When a drum is beaten, its membrane vibrates and produces sound. The vibrating object acts as the source of sound.

Production of Compressions and Rarefactions:

Consider a vibrating tuning fork as the source of sound.

- When the prong of the tuning fork moves outward (forward), it pushes the air particles in front of it. These particles get crowded together, forming a region of high pressure called compression (C).
- When the prong moves inward (backward), it creates a region where air particles are spread apart, forming a region of low pressure called rarefaction (R).
- As the tuning fork continues to vibrate, a series of compressions and rarefactions are produced alternately in the air.

Diagram (description):
$$\text{Source} \rightarrow \underbrace{||||}_{C} \quad \underbrace{\quad\quad}_{R} \quad \underbrace{||||}_{C} \quad \underbrace{\quad\quad}_{R} \quad \underbrace{||||}_{C} \rightarrow$$

Where $C$ = Compression (region of high pressure, particles close together) and $R$ = Rarefaction (region of low pressure, particles spread apart).

These compressions and rarefactions propagate through the air as a longitudinal sound wave.

Sound wave is called a longitudinal wave because the particles of the medium vibrate in the same direction as the direction of propagation of the wave.

Explanation:
When a sound wave travels through air, the air particles are displaced back and forth along the direction in which the wave is moving. This creates alternating regions of compressions (high pressure) and rarefactions (low pressure) along the direction of wave propagation.

Since the direction of particle vibration is parallel (along the same line) to the direction of wave propagation, sound waves are classified as longitudinal waves.

This is in contrast to transverse waves (like light waves), where particle vibration is perpendicular to the direction of wave propagation.

The characteristic of sound that helps us identify a friend by his voice in a dark room is quality (or timbre) of sound.

Explanation:
Even if two sounds have the same pitch (frequency) and the same loudness (amplitude), they can differ in quality. Quality of sound depends on the waveform of the sound, which is determined by the number and relative intensities of the overtones (harmonics) present in the sound.

Every person has a unique vocal structure, so the sound produced by each person has a unique waveform (quality/timbre). This unique quality allows us to distinguish one person's voice from another even in darkness.

Given/Known:
- Flash (light) and thunder (sound) are produced simultaneously during lightning.
- Speed of light $= 3 \times 10^8 \ \text{m s}^{-1}$
- Speed of sound in air $\approx 344 \ \text{m s}^{-1}$

Reason:
The speed of light is enormously greater than the speed of sound. Light travels at $3 \times 10^8 \ \text{m s}^{-1}$, while sound travels at only about $344 \ \text{m s}^{-1}$ in air.

Therefore, light from the flash reaches our eyes almost instantaneously, while sound (thunder) takes a much longer time to travel the same distance. This is why we see the flash first and hear the thunder a few seconds later.

Conclusion: The time gap between seeing the flash and hearing the thunder is due to the vast difference in the speeds of light and sound.

Given:
- Speed of sound in air, $v = 344 \ \text{m s}^{-1}$
- Frequency $f_1 = 20 \ \text{Hz}$
- Frequency $f_2 = 20 \ \text{kHz} = 20{,}000 \ \text{Hz}$

Formula:
$$\lambda = \frac{v}{f}$$

For $f_1 = 20 \ \text{Hz}$:
$$\lambda_1 = \frac{344}{20} = 17.2 \ \text{m}$$

For $f_2 = 20{,}000 \ \text{Hz}$:
$$\lambda_2 = \frac{344}{20{,}000} = 0.0172 \ \text{m} = 1.72 \ \text{cm}$$

Result:
- Wavelength corresponding to 20 Hz $= 17.2 \ \text{m}$
- Wavelength corresponding to 20 kHz $= 0.0172 \ \text{m}$ (or $1.72 \ \text{cm}$)

Given:
- Speed of sound in air, $v_{\text{air}} = 346 \ \text{m s}^{-1}$ (standard value used in NCERT)
- Speed of sound in aluminium, $v_{\text{Al}} = 6420 \ \text{m s}^{-1}$ (standard value from NCERT table)
- Let the length of the aluminium rod $= d$

Formula:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Time taken by sound through air:
$$t_{\text{air}} = \frac{d}{v_{\text{air}}} = \frac{d}{346}$$

Time taken by sound through aluminium:
$$t_{\text{Al}} = \frac{d}{v_{\text{Al}}} = \frac{d}{6420}$$

Ratio:
$$\frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{d/346}{d/6420} = \frac{6420}{346} \approx 18.55 \approx 18.5$$

$$\boxed{\frac{t_{\text{air}}}{t_{\text{Al}}} \approx 18.5 : 1}$$

This means sound takes about 18.5 times longer to travel through air than through aluminium for the same distance.

Given:
- Frequency of sound, $f = 100 \ \text{Hz}$
- Time $= 1 \ \text{minute} = 60 \ \text{seconds}$

Concept:
Frequency is defined as the number of vibrations (oscillations) per second.
$$f = \frac{\text{Number of vibrations}}{\text{Time (in seconds)}}$$

Calculation:
$$\text{Number of vibrations in 1 minute} = f \times t = 100 \ \text{Hz} \times 60 \ \text{s}$$
$$= 6000 \ \text{vibrations}$$

Answer: The source vibrates 6000 times in one minute.

Yes, sound follows the same laws of reflection as light does.

Laws of Reflection of Sound:

1. First Law: The angle of incidence of the sound wave is equal to the angle of reflection.
$$\angle i = \angle r$$

2. Second Law: The incident sound wave, the reflected sound wave, and the normal to the reflecting surface at the point of incidence — all lie in the same plane.

Explanation:
Just like light, when sound waves strike a hard surface (like a wall, cliff, or building), they bounce back. The direction of the reflected sound wave obeys the same geometrical rules as reflected light. This is why we hear echoes — the reflected sound follows the same laws as reflected light.

Practical evidence: The phenomenon of echo and the working of a megaphone, stethoscope, and soundboards all demonstrate that sound obeys the laws of reflection.

Yes, the echo will be heard sooner on a hotter day (but it will still be heard).

Explanation:
- The speed of sound increases with an increase in temperature. On a hotter day, the speed of sound in air is greater.
- The minimum distance required to hear an echo is:
$$d = \frac{v \times t}{2}$$
where $t = 0.1 \ \text{s}$ (minimum time for the human ear to distinguish two sounds) and $v$ is the speed of sound.
- On a hotter day, since $v$ is larger, the minimum distance required to hear an echo also increases.
- However, if the distance between the source and the reflecting surface remains the same and was sufficient to produce an echo at normal temperature, then on a hotter day the echo will return faster (in less time) because sound travels faster.
- If the distance was just barely sufficient at normal temperature, it is possible that on a hotter day the echo returns so quickly (less than 0.1 s) that it may not be distinguishable as a separate echo.

Conclusion: On a hotter day, sound travels faster, so the echo is heard sooner. Whether it is perceived as a distinct echo depends on whether the time gap remains at least 0.1 seconds.

Two practical applications of reflection of sound waves:

1. Megaphone / Loudspeaker / Horn:
Megaphones, horns, and musical instruments like trumpets are designed to send sound in a particular direction without spreading it in all directions. The sound is reflected repeatedly from the walls of the horn/tube and directed towards the audience. This is based on the reflection of sound.

2. Stethoscope:
A stethoscope is a medical instrument used by doctors to listen to sounds produced within the body (e.g., heartbeat, breathing). The sound of the heartbeat reaches the doctor's ears by multiple reflections of sound waves through the rubber tube of the stethoscope. This works on the principle of reflection of sound.

*(Other valid applications include: SONAR, soundboards in auditoriums, megaphones, and echoes used in measuring distances.)*

Given:
- Height of tower, $h = 500 \ \text{m}$
- Acceleration due to gravity, $g = 10 \ \text{m s}^{-2}$
- Speed of sound, $v_s = 340 \ \text{m s}^{-1}$
- Initial velocity of stone, $u = 0$

Step 1: Time taken by the stone to fall to the base ($t_1$)

Using the equation of motion:
$$h = ut_1 + \frac{1}{2}g t_1^2$$
$$500 = 0 + \frac{1}{2} \times 10 \times t_1^2$$
$$500 = 5 t_1^2$$
$$t_1^2 = 100$$
$$t_1 = 10 \ \text{s}$$

Step 2: Time taken by sound to travel from base to top ($t_2$)

$$t_2 = \frac{h}{v_s} = \frac{500}{340} \approx 1.47 \ \text{s}$$

Step 3: Total time after which splash is heard at the top

$$t = t_1 + t_2 = 10 + 1.47 = 11.47 \ \text{s}$$

Answer: The splash will be heard at the top of the tower after approximately $11.47 \ \text{s}$ of dropping the stone.

Given:
- Speed of sound, $v = 339 \ \text{m s}^{-1}$
- Wavelength, $\lambda = 1.5 \ \text{cm} = 1.5 \times 10^{-2} \ \text{m} = 0.015 \ \text{m}$

Formula:
$$v = f \lambda \implies f = \frac{v}{\lambda}$$

Calculation:
$$f = \frac{339}{0.015} = 22{,}600 \ \text{Hz} = 22.6 \ \text{kHz}$$

Audibility Check:
The audible range for human beings is $20 \ \text{Hz}$ to $20{,}000 \ \text{Hz}$ (20 kHz).

Since f = 22{,}600 \ \text{Hz} > 20{,}000 \ \text{Hz}, this frequency is above the audible range.

Answer: The frequency of the wave is $22{,}600 \ \text{Hz}$ (22.6 kHz). It will not be audible to human beings as it falls in the ultrasonic range.

Reverberation:
Reverberation is the persistence of sound in a large enclosed space (like a big hall or auditorium) due to repeated reflections of sound from the walls, ceiling, and floor even after the source of sound has stopped. The sound appears to linger or "echo" within the room.

For reverberation to be perceived, the time gap between the original sound and the reflected sound must be less than $\frac{1}{10}$ second (0.1 s), so the sounds overlap and seem to persist.

How to reduce reverberation:

1. Sound-absorbing materials: The walls, ceiling, and floor of auditoriums and concert halls are covered with sound-absorbing materials such as compressed fibreboard, rough plaster, draperies, and carpets. These materials absorb sound energy and reduce reflections.

2. Seats and audience: Seats are often made of materials that absorb sound. The presence of an audience also helps absorb sound.

3. Curtains and soft furnishings: Heavy curtains, carpets, and upholstered furniture in rooms absorb sound and reduce reverberation.

4. Acoustic tiles: Special acoustic tiles are used on walls and ceilings to absorb sound and minimise multiple reflections.

Loudness of Sound:
Loudness is a physiological (subjective) response of the ear to the intensity of sound. It is the measure of the sound energy reaching the ear per unit time. Louder sounds carry more energy.

The unit of loudness is the decibel (dB).

Factors on which loudness depends:

1. Amplitude of vibration: Loudness is directly proportional to the square of the amplitude of the vibrating source.
$$\text{Loudness} \propto A^2$$
Greater the amplitude of vibration, louder is the sound.

2. Distance from the source: As the distance from the source increases, the loudness decreases because the sound energy spreads over a larger area.

3. Surface area of the vibrating body: A larger vibrating surface sets more air particles into vibration, producing a louder sound.

4. Density of the medium: Sound is louder in denser media because more particles vibrate per unit volume.

Summary: Loudness primarily depends on the amplitude of the sound wave — greater amplitude means greater loudness.

Ultrasound for Cleaning:

Ultrasound waves are used to clean objects that are in hard-to-reach places, such as spiral tubes, odd-shaped parts, and electronic components.

Process:
1. The objects to be cleaned are placed in a cleaning solution (water or a suitable solvent) in a tank.
2. Ultrasonic waves (high-frequency sound waves, frequency > 20{,}000 \ \text{Hz}) are passed through the solution.
3. The high-frequency vibrations of the ultrasonic waves cause the dirt, grease, and other contaminants attached to the objects to vibrate vigorously.
4. Due to this vigorous vibration, the dirt and grease particles get detached from the surface of the object and fall into the cleaning solution.
5. The object is thus thoroughly cleaned without any physical scrubbing.

Advantage: This method can clean delicate and intricate objects without causing any mechanical damage to them.

Detection of Defects in Metal Blocks Using Ultrasound:

Metal blocks used in construction and machinery may have internal cracks, holes, or other defects that are not visible from the outside. Ultrasound is used to detect such defects.

Method:
1. Ultrasonic waves (frequency > 20{,}000 \ \text{Hz}) are directed into the metal block from one side using an ultrasonic transmitter.
2. These waves travel through the metal block.
3. If there is no defect, the ultrasonic waves pass straight through the block and are detected by a detector on the other side.
4. If there is a crack, hole, or defect inside the metal block, the ultrasonic waves are reflected back from the defect (because the defect acts as a boundary between two different media).
5. The reflected waves are detected by a detector, and the location and size of the defect can be determined from the time taken and the pattern of the reflected waves.

Principle used: Reflection of ultrasound at the boundary of a defect (crack or cavity).

Advantage: Ultrasound can penetrate deep into metal blocks and detect internal defects without cutting or damaging the metal, making it a non-destructive testing (NDT) method.