NCERT Class 9 Science Solutions: Sound
Question:
Why are the ceilings of concert halls curved?
Concept in a Minute:
Sound reflection, specifically the principle of reflection where the angle of incidence equals the angle of reflection. Echoes and reverberation are also relevant.
Explanation:
The ceilings of concert halls are curved to ensure that sound waves are reflected evenly and directed towards the audience. Imagine a sound wave originating from the stage; if the ceiling were flat, the sound would reflect directly upwards or in unpredictable directions, leading to poor acoustics and potential echoes in some areas and a lack of sound in others.
By making the ceiling curved, it acts like a series of curved surfaces (parabolic or elliptical shapes are often used). These shapes are designed to reflect sound waves from the stage in a controlled manner. The curvature ensures that sound waves hitting the ceiling are reflected downwards and outwards, spreading evenly across the entire seating area. This uniform distribution of sound helps to:
1. Prevent Echoes and Dead Spots: The curved surface breaks up and disperses sound waves, preventing them from bouncing directly back to create distracting echoes. It also ensures that no areas of the hall are “dead spots” where sound is significantly weaker.
2. Enhance Sound Quality: By reflecting sound waves efficiently and evenly, the curvature contributes to a richer, fuller sound experience for all members of the audience, regardless of their seating position. It ensures that the sound reaches the listener directly from the source (stage) and also indirectly via reflections from the ceiling, creating a sense of warmth and presence.
3. Improve Intelligibility: The controlled reflection of sound helps to maintain the clarity and intelligibility of speech and music.
In essence, the curved ceiling acts as a large-scale acoustic reflector, optimizing the way sound travels and reaches the audience for the best possible listening experience.
Question:
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Concept in a Minute:
The speed of sound depends on the medium through which it travels. Sound travels faster in denser media. Denser media have more particles packed closely together, allowing vibrations to be transmitted more efficiently.
Explanation:
Sound is a mechanical wave, meaning it requires a medium to propagate. The speed of sound in a medium is determined by the medium’s elasticity and density. Generally, sound travels faster in solids than in liquids, and faster in liquids than in gases. This is because solids are generally denser and more rigid (elastic) than liquids, and liquids are denser and more rigid than gases. Iron is a solid, water is a liquid, and air is a gas. At a particular temperature, iron is the densest and most rigid of the three media. Therefore, sound will travel the fastest in iron.
Detailed Solution:
1. Understand the nature of sound: Sound is a mechanical wave that requires a medium for its propagation.
2. Recall factors affecting the speed of sound: The speed of sound depends on the properties of the medium, primarily its elasticity (rigidity) and density.
3. Compare the states of matter: The question provides three media: air (gas), water (liquid), and iron (solid).
4. Relate state of matter to density and rigidity: Generally, solids are denser and more rigid than liquids, and liquids are denser and more rigid than gases.
5. Apply the concept to the given media:
– Air is a gas, which is least dense and least rigid.
– Water is a liquid, which is denser and more rigid than air.
– Iron is a solid, which is significantly denser and more rigid than both water and air.
6. Conclude the speed of sound: Since sound travels faster in denser and more rigid media, it will travel fastest in iron.
Final Answer: In iron, sound travels the fastest at a particular temperature.
Question:
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Concept in a Minute:
The concept of echo production relies on the reflection of sound waves from a surface. The key factors influencing the perception of an echo are the speed of sound and the time it takes for the sound to travel to the reflecting surface and back to the listener. The speed of sound is affected by the medium through which it travels, particularly its temperature.
Explanation:
Yes, you are more likely to hear an echo on a hotter day. An echo is heard when the reflected sound wave reaches the listener at least 0.1 seconds after the original sound. The speed of sound in air increases with an increase in temperature. On a hotter day, the air is warmer, and therefore, sound travels faster. Since the distance to the reflecting surface remains the same, a faster speed of sound means the sound wave will cover the distance to the object and back in a shorter amount of time. If this time is less than 0.1 seconds, you won’t perceive it as a distinct echo. However, if the speed of sound is sufficiently high (on a hotter day), the time taken for the sound to travel to the object and return might still be longer than the minimum required for a distinct echo, or it might be just at the threshold where the echo is noticeable. Conversely, on a colder day, sound travels slower, and it takes longer for the reflected sound to return, making echoes more easily distinguishable. Therefore, while the condition for echo is distance and time, the *perception* of a distinct echo is influenced by the speed of sound, which is higher on hotter days. This means the time taken for the round trip is less, and the echo will be heard sooner after the original sound. This temporal separation is crucial for distinguishing the echo from the original sound.
Question:
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Concept in a Minute:
The key concept here is the definition of frequency and its relationship to the time period of a wave. Frequency is the number of cycles (compressions and rarefactions for sound) per second. The time period is the time taken for one complete cycle. They are reciprocally related. The distance of the listener from the source is irrelevant to the time interval between successive compressions.
Explanation:
The question asks for the time interval between successive compressions from the source. This time interval is the period of the sound wave. The frequency of the sound tone is given as 500 Hz.
Frequency (f) is defined as the number of complete cycles of a wave that pass a point per second. The unit of frequency is Hertz (Hz), where 1 Hz = 1 cycle/second.
The time period (T) is the time taken for one complete cycle of the wave.
The relationship between frequency and time period is given by:
T = 1/f
Given:
Frequency (f) = 500 Hz
Therefore, the time interval between successive compressions (which is the time period) is:
T = 1/500 seconds
T = 0.002 seconds
The distance of the listener from the source (450 m) is extraneous information and is not needed to solve this problem.
Detailed Steps:
1. Identify the given information: Frequency of the sound tone = 500 Hz.
2. Understand what is being asked: The time interval between successive compressions from the source. This corresponds to the time period of the sound wave.
3. Recall the relationship between frequency and time period: T = 1/f.
4. Substitute the given frequency into the formula.
5. Calculate the time period.
6. State the final answer with appropriate units.
Question:
How can reverberation be reduced?
Concept in a Minute:
Reverberation is the persistence of sound in a space after the original sound has stopped, caused by multiple reflections of sound waves off surfaces. Reducing reverberation involves minimizing these reflections.
Explanation:
Reverberation can be reduced by using sound-absorbing materials. These materials convert sound energy into heat energy, thus preventing the sound waves from reflecting back into the room. Examples of such materials include:
1. Soft furnishings: Curtains, carpets, upholstered furniture, and tapestries are effective at absorbing sound.
2. Acoustic panels: These are specially designed panels made from porous materials like fiberglass or mineral wool, often covered with fabric, and are placed on walls and ceilings.
3. Perforated materials: Materials with small holes, like perforated ceiling tiles or wooden panels with many small perforations, can absorb sound because the sound waves penetrate the holes and get trapped within the material.
4. Open spaces: Large, open spaces without many hard surfaces tend to have less reverberation than enclosed spaces with hard walls. While not a direct reduction method in an existing space, it’s a principle to consider in design.
5. Strategic placement of sound sources and listeners: In some cases, repositioning sound sources or listeners can alter the path of reflections and reduce the perceived reverberation.
In essence, the key to reducing reverberation is to introduce materials that absorb sound rather than reflecting it.
Question:
What are wavelength, frequency, time period and amplitude of a sound wave?
Concept in a Minute:
Sound waves are mechanical waves that travel through a medium by creating compressions and rarefactions. Understanding these properties helps describe the characteristics of the wave, such as its pitch and loudness.
Explanation:
Wavelength: The wavelength of a sound wave is the distance between two consecutive compressions or two consecutive rarefactions. It is typically denoted by the Greek letter lambda ($\lambda$) and is measured in meters (m).
Frequency: The frequency of a sound wave is the number of complete oscillations or cycles that pass a given point in one second. It determines the pitch of the sound. Higher frequency means higher pitch. It is denoted by ‘f’ and measured in Hertz (Hz). 1 Hz is equal to one cycle per second.
Time Period: The time period of a sound wave is the time taken for one complete oscillation or cycle to pass a given point. It is the reciprocal of frequency and is denoted by ‘T’. It is measured in seconds (s). The relationship between time period and frequency is T = 1/f.
Amplitude: The amplitude of a sound wave is the maximum displacement or variation of the particles of the medium from their mean position. It is related to the intensity or loudness of the sound. A larger amplitude corresponds to a louder sound. It is typically denoted by ‘A’ and measured in units of displacement, such as meters (m) or Pascals (Pa) for pressure variations.
Question:
An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m s−1?
Concept in a Minute:
This question involves the concept of echoes. An echo is a reflection of sound. To calculate the distance, we need to understand that the sound travels to the reflecting surface and then back to the source. The total time given is for this round trip. We will use the formula: distance = speed × time.
Explanation:
The problem states that an echo returned in 3 seconds. This means that the sound traveled from the source to the reflecting surface and then back to the source, covering the distance twice, in 3 seconds.
The speed of sound is given as 342 m s−1.
Let ‘d’ be the distance of the reflecting surface from the source.
The total distance traveled by the sound to form an echo is 2d (distance to the surface + distance back from the surface).
The time taken for this round trip is given as t = 3 s.
The speed of sound is given as v = 342 m s−1.
We know the relationship between distance, speed, and time:
distance = speed × time
In this case, the total distance is 2d, so:
2d = v × t
Now, we can substitute the given values into the equation:
2d = 342 m s−1 × 3 s
2d = 1026 m
To find the distance ‘d’ (the distance of the reflecting surface from the source), we need to divide the total distance by 2:
d = 1026 m / 2
d = 513 m
Therefore, the distance of the reflecting surface from the source is 513 meters.
The final answer is $\boxed{513 m}$.
Question:
What is loudness of sound? What factors does it depend on?
Concept in a Minute:
The loudness of sound is related to its intensity. Intensity is the amount of energy a sound wave carries per unit area per unit time. Factors that affect loudness include the amplitude of the sound wave and the distance from the source.
Explanation:
Loudness of sound is the subjective perception of the intensity of a sound. It is how strong or weak a sound appears to our ears. The intensity of a sound wave is determined by the square of its amplitude. Amplitude, in turn, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position.
Factors influencing the loudness of sound:
1. Amplitude of the sound wave: A sound wave with a larger amplitude carries more energy and therefore has higher intensity, resulting in a louder sound. Conversely, a smaller amplitude means less energy and a softer sound.
2. Distance from the source: The intensity of a sound wave decreases as the distance from the source increases. This is because the sound energy spreads out over a larger area as it travels. Therefore, a sound will be perceived as louder when you are closer to the source and softer when you are farther away.
In summary, loudness is primarily dependent on the amplitude of the sound wave (which relates to the energy of the vibration) and the distance from the sound source.
Question:
Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Concept in a Minute:
The key concept needed to solve this question is the characteristic of sound that allows for the identification of a person’s voice. This characteristic is related to the unique quality of sound produced by each individual’s vocal cords and resonating cavities.
Explanation:
The characteristic of sound that helps you identify your friend by his voice is called timbre. Timbre, also known as tone quality or tone color, is what distinguishes different types of sound production, such as voices and musical instruments, even when they are producing the same pitch and loudness. It is determined by the complex mixture of overtones present in the sound wave, as well as the attack and decay of the sound. Each person’s voice has a unique timbre due to the specific size and shape of their vocal cords, larynx, and vocal tract, which influences how sound waves are produced and resonated. In a dark room, when you can’t see who is speaking, it is this unique timbre of your friend’s voice that allows you to recognize them among others. While pitch relates to how high or low a sound is, and loudness relates to how strong it is, timbre is the distinct quality that makes voices different.
Question:
Does sound follow the same laws of reflection as light does? Explain.
Concept in a Minute:
The fundamental principles of reflection apply to both sound and light waves. These principles involve the angle of incidence, angle of reflection, and the behavior of waves when they encounter a surface.
Explanation:
Yes, sound follows the same laws of reflection as light does. The laws of reflection state that:
1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the reflected ray, and the normal to the surface at the point of incidence all lie in the same plane.
When a sound wave strikes a surface, it bounces back. The angle at which the sound wave hits the surface (angle of incidence) is precisely the same as the angle at which it bounces off (angle of reflection). Similarly, the path of the incoming sound wave (incident ray), the path of the reflected sound wave (reflected ray), and an imaginary line perpendicular to the surface at the point where the sound hits (normal) all exist on the same plane. This is why we can observe phenomena like echoes, which are a direct result of sound reflection obeying these laws. Just as light reflects off mirrors to form images, sound reflects off surfaces to create echoes.
Question:
A sound wave travels at a speed of 339 m s−1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Concept in a Minute:
The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by the formula v = fλ. Audibility of a sound wave depends on its frequency falling within the human hearing range, typically between 20 Hz and 20,000 Hz.
Explanation:
The problem provides the speed of the sound wave (v = 339 m/s) and its wavelength (λ = 1.5 cm). To find the frequency (f), we can rearrange the wave equation: f = v/λ.
First, we need to ensure that the units are consistent. The speed is given in meters per second (m/s), but the wavelength is in centimeters (cm). We must convert the wavelength to meters.
1.5 cm = 1.5 / 100 meters = 0.015 meters.
Now, we can plug the values into the formula:
f = 339 m/s / 0.015 m
f = 22600 Hz
To determine if the sound is audible, we compare its frequency to the human hearing range (20 Hz to 20,000 Hz). The calculated frequency is 22,600 Hz.
Since 22,600 Hz is greater than 20,000 Hz, the sound wave is above the upper limit of human hearing. Therefore, it will not be audible to humans.
Final Answer: The frequency of the wave is 22600 Hz. It will not be audible because it is above the human hearing range.
Question:
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s−1.
Concept in a Minute:
The relationship between the speed of a wave, its frequency, and its wavelength is given by the wave equation: speed = frequency × wavelength. This equation is fundamental to understanding wave phenomena.
Explanation:
The question asks us to find the wavelengths of sound waves in air corresponding to the lower and upper limits of a human hearing range. We are given the frequencies and the speed of sound in air.
The fundamental relationship between speed (v), frequency (f), and wavelength (λ) of a wave is:
v = fλ
We can rearrange this equation to solve for wavelength:
λ = v / f
We are given:
Speed of sound in air, v = 344 m s⁻¹
For the lower frequency limit:
Frequency, f₁ = 20 Hz
Wavelength, λ₁ = v / f₁
λ₁ = 344 m s⁻¹ / 20 Hz
λ₁ = 17.2 m
For the upper frequency limit:
Frequency, f₂ = 20 kHz
First, convert kHz to Hz: 20 kHz = 20 × 10³ Hz = 20,000 Hz
Wavelength, λ₂ = v / f₂
λ₂ = 344 m s⁻¹ / 20,000 Hz
λ₂ = 0.0172 m
Therefore, the typical wavelengths of sound waves in air corresponding to the hearing range from 20 Hz to 20 kHz are 17.2 m and 0.0172 m, respectively.
Question:
How are the wavelength and frequency of a sound wave related to its speed?
Concept in a Minute:
The fundamental relationship between wave speed, wavelength, and frequency. Wave speed is directly proportional to both wavelength and frequency.
Explanation:
The speed of a sound wave (v) is determined by the product of its wavelength (λ) and its frequency (f). This relationship is expressed by the formula:
v = λf
Where:
v = speed of the sound wave (usually measured in meters per second, m/s)
λ = wavelength of the sound wave (usually measured in meters, m)
f = frequency of the sound wave (usually measured in Hertz, Hz, which is cycles per second)
This means that if the speed of sound is constant in a medium:
– If the frequency of the sound wave increases, its wavelength must decrease to keep the speed constant.
– If the wavelength of the sound wave increases, its frequency must decrease to keep the speed constant.
For example, in air at room temperature, the speed of sound is approximately 343 m/s. A high-pitched sound (high frequency) will have a short wavelength, while a low-pitched sound (low frequency) will have a long wavelength.
Question:
What is reverberation?
Concept in a Minute:
Sound waves reflect off surfaces. When these reflections are numerous and persist for a noticeable duration, they create reverberation. This is related to the persistence of hearing, where the brain needs a short time to distinguish between separate sounds.
Explanation:
Reverberation is the persistence of sound in a space after the original sound has stopped. It is caused by multiple reflections of sound waves bouncing off the walls, ceiling, floor, and other objects in an enclosed space. When these reflected sound waves reach the listener’s ear in quick succession and at varying intensities and times, they combine with the direct sound and with each other. This causes the sound to appear to linger or echo. If the reflections are too frequent and intense, they can blur the original sound, making speech or music unclear. This phenomenon is particularly noticeable in large, empty rooms or auditoriums with hard surfaces.
Question:
How does the sound produced by a vibrating object in a medium reach your ear?
Concept in a Minute:
Sound is produced by vibrations. These vibrations create disturbances in the medium. These disturbances travel as waves. Our ears detect these waves.
Explanation:
When an object vibrates, it pushes and pulls on the particles of the medium surrounding it (like air, water, or solids). This pushing action compresses the particles together, creating a region of high pressure called a compression. The pulling action stretches the particles apart, creating a region of low pressure called a rarefaction. These compressions and rarefactions travel outwards from the vibrating object as a longitudinal wave. This wave is what we call sound. When this sound wave reaches your ear, it causes your eardrum to vibrate. These vibrations are then transmitted through the small bones in your middle ear and converted into electrical signals by the inner ear, which are then sent to your brain, where they are interpreted as sound.
Question:
What is the range of frequencies associated with ultrasound?
Concept in a Minute:
The question is about ultrasound frequencies. Ultrasound refers to sound waves with frequencies higher than the upper audible limit of human hearing. The human hearing range is generally considered to be from 20 Hz to 20 kHz. Therefore, ultrasound frequencies are above 20 kHz.
Explanation:
The range of frequencies associated with ultrasound is defined as sound waves that have a frequency greater than the upper limit of human hearing. The typical upper limit of human hearing is approximately 20,000 Hertz (20 kHz). Therefore, ultrasound frequencies are frequencies above 20 kHz. In practical applications and scientific contexts, ultrasound frequencies are often considered to be in the range of 20 kHz to several gigahertz (GHz). However, the defining characteristic is that they are beyond the audible spectrum for humans.
Range of frequencies associated with ultrasound: Above 20 kHz.
Question:
Which wave property determines pitch?
Concept in a Minute:
Sound waves are characterized by several properties like amplitude, frequency, and wavelength. Pitch is a perceptual characteristic of sound that relates to how high or low it sounds. Frequency is the physical property of a sound wave that directly corresponds to our perception of pitch.
Explanation:
Pitch is determined by the frequency of the sound wave. Frequency is defined as the number of complete cycles or oscillations a wave makes in one second. It is measured in Hertz (Hz), where 1 Hz is equal to one cycle per second. A higher frequency corresponds to a higher pitch (a shriller sound), while a lower frequency corresponds to a lower pitch (a deeper sound). For example, a soprano singer produces sound waves with higher frequencies than a bass singer. Therefore, frequency is the wave property that determines pitch.
Question:
Why are sound waves called mechanical waves?
Concept in a Minute:
Mechanical waves are waves that require a medium to propagate. They transfer energy through the vibration of particles within that medium. Sound waves are a type of wave that needs a substance like air, water, or solids to travel.
Explanation:
Sound waves are called mechanical waves because they need a material medium to travel from one place to another. When a sound is produced, it causes the particles of the medium (like air molecules) to vibrate. These vibrations are passed on from one particle to the next, creating a wave of compressions and rarefactions that propagates through the medium. Without such a medium, sound cannot travel. For instance, sound cannot travel in a vacuum because there are no particles to vibrate. This dependency on a medium for propagation is the defining characteristic of mechanical waves.
Question:
Give two practical applications of reflection of sound waves.
Concept in a Minute:
Reflection of sound waves is the phenomenon where sound waves bounce back after striking a surface. This principle is used in various applications.
Explanation:
Two practical applications of reflection of sound waves are:
1. Megaphone/Loudhailer: A megaphone is designed to amplify sound. Its conical shape reflects the sound waves generated by the speaker in a specific direction, preventing them from dispersing in all directions. This concentrated sound reaches the audience more effectively, making it louder.
2. Sounding Board: A sounding board is a flat, hard surface placed behind a musical instrument (like a piano or violin) or a speaker. It reflects the sound waves produced by the instrument or speaker. This reflection adds to the overall loudness and richness of the sound, making it more resonant and audible.
Question:
Which wave property determines loudness?
Concept in a Minute:
Sound waves are characterized by certain properties like amplitude, frequency, and wavelength. Amplitude is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. Loudness is how we perceive the intensity of sound.
Explanation:
The loudness of a sound is determined by its amplitude. A sound wave with a larger amplitude carries more energy, which our ears perceive as a louder sound. Conversely, a sound wave with a smaller amplitude has less energy and is perceived as a softer sound. Frequency determines the pitch of the sound (high frequency means high pitch, low frequency means low pitch). Wavelength is related to both frequency and the speed of the wave. Therefore, amplitude is the wave property that determines loudness.
Question:
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Concept in a Minute:
Frequency is the number of vibrations or oscillations a source makes per second. The relationship between frequency, time, and the total number of vibrations is key.
Explanation:
The frequency of the sound source is given as 100 Hz. This means the source vibrates 100 times every second.
We need to find out how many times it vibrates in a minute.
First, we need to convert the time from minutes to seconds, because frequency is defined in vibrations per second.
There are 60 seconds in 1 minute.
So, if the source vibrates 100 times in 1 second, then in 60 seconds (which is 1 minute), it will vibrate:
Total vibrations = Frequency × Time in seconds
Total vibrations = 100 vibrations/second × 60 seconds
Total vibrations = 6000 vibrations.
Therefore, the source vibrates 6000 times in a minute.
Question:
How is ultrasound used for cleaning?
Concept in a Minute:
Cavitation, high frequency sound waves, mechanical action, removal of dirt.
Explanation:
Ultrasound cleaning utilizes the phenomenon of cavitation. High-frequency sound waves are passed through a cleaning liquid. These sound waves create rapidly forming and collapsing microscopic bubbles in the liquid. This process, called cavitation, generates intense localized forces. When these bubbles collapse near the surface of an object being cleaned, they produce tiny jets of liquid that dislodge dirt, grease, and other contaminants. The high frequency ensures that these bubbles form and collapse millions of times per second, providing a vigorous yet gentle scrubbing action that reaches even intricate parts and crevices, effectively cleaning the object.
Question:
Why is sound wave called a longitudinal wave?
Concept in a Minute:
Longitudinal waves involve particle displacement parallel to the wave’s direction of propagation. This creates compressions and rarefactions.
Explanation:
A sound wave is called a longitudinal wave because the particles of the medium (like air molecules) vibrate back and forth in the same direction as the wave is traveling. Imagine pushing and pulling a spring; the coils compress and stretch in the direction you are pushing and pulling. Similarly, in a sound wave, the air molecules are pushed together (forming a compression) and then spread apart (forming a rarefaction). This back-and-forth motion of the particles, parallel to the direction the sound energy is moving, is the defining characteristic of a longitudinal wave.
Question:
What is sound and how is it produced?
Concept in a Minute:
Sound is a form of energy that travels as waves. It is produced by vibrations.
Explanation:
Sound is a vibration that travels through a medium (like air, water, or solids) as a wave. When an object vibrates, it causes the particles of the medium around it to vibrate as well. These vibrations are passed from one particle to another, creating a wave of compression and rarefaction that travels outwards from the source. This wave is what we perceive as sound.
For example, when you speak, your vocal cords vibrate, causing the air molecules around them to vibrate. These vibrations travel to your ears, and your brain interprets them as sound. Similarly, when a drum is struck, the drumhead vibrates, producing sound waves.
Question:
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Concept in a Minute:
The relationship between the speed of a wave, its frequency, and its wavelength is fundamental in wave physics. This relationship is expressed by the wave equation: speed = frequency × wavelength.
Explanation:
We are given the frequency (f) of the sound wave as 220 Hz and the speed (v) of the sound wave in the medium as 440 m/s. We need to calculate the wavelength (λ) of the sound wave.
The formula that relates speed, frequency, and wavelength is:
v = f × λ
To find the wavelength (λ), we can rearrange the formula:
λ = v / f
Now, substitute the given values into the formula:
λ = 440 m/s / 220 Hz
Calculate the result:
λ = 2 meters
Therefore, the wavelength of the sound wave is 2 meters.
Question:
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Concept in a Minute:
Speed of light vs. Speed of sound
Explanation:
Flash and thunder are produced at the same time during a lightning strike. However, we see the flash before we hear the thunder because light travels much, much faster than sound. The speed of light is approximately 300,000 kilometers per second, while the speed of sound in air is about 343 meters per second. Therefore, the light from the flash reaches our eyes almost instantaneously, while the sound waves of the thunder take a noticeable amount of time to travel the distance from the lightning strike to our ears. This difference in speed causes the delay between seeing the flash and hearing the thunder.
Question:
What is the audible range of the average human ear?
Concept in a Minute:
The audible range of human hearing refers to the spectrum of sound frequencies that a typical human can perceive. This range is typically expressed in Hertz (Hz), which measures the number of sound wave cycles per second.
Explanation:
The audible range of the average human ear is generally considered to be from approximately 20 Hertz (Hz) to 20,000 Hertz (20 kHz). Sounds with frequencies below 20 Hz are called infrasound, and sounds with frequencies above 20,000 Hz are called ultrasound. While this is the typical range, it can vary slightly between individuals and tends to decrease with age, particularly at the higher frequency end.
Question:
Explain how sound is produced by your school bell.
Concept in a Minute:
Vibrations and Sound Production. Sound is produced when an object vibrates. These vibrations travel through a medium (like air) as waves and reach our ears, where they are interpreted as sound.
Explanation:
When you strike a school bell, the metal of the bell is made to vibrate. These rapid back-and-forth movements of the bell’s material create disturbances in the surrounding air. These disturbances travel outwards as sound waves. When these sound waves reach your ears, they cause your eardrums to vibrate, and your brain interprets these vibrations as the sound of the bell.
Question:
What is the range of frequencies associated with infrasound?
Concept in a Minute:
Sound waves are classified by their frequency. The human ear can typically hear sounds within a certain frequency range. Sounds with frequencies below this audible range are called infrasound, and sounds with frequencies above this audible range are called ultrasound.
Explanation:
The range of frequencies associated with infrasound is below the lower limit of human hearing. The typical range of human hearing is from approximately 20 Hertz (Hz) to 20,000 Hertz (20 kHz). Therefore, infrasound refers to sound waves with frequencies less than 20 Hz.
Question:
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Concept in a Minute:
Sound requires a medium for propagation.
Explanation:
No, you will not be able to hear any sound produced by your friend on the moon. Sound waves are mechanical waves, meaning they need a medium (like air, water, or solids) to travel. The moon has virtually no atmosphere, which means there is no medium for sound to travel through from your friend to you. Therefore, even if your friend makes a loud noise, it will not be audible.
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