Mix Example - SOUND Questions in English

(A) note is defined as a sound produced by a mixture of several frequencies. It is characterized by being pleasant to hear,unlike noise,which is typically unpleasant.

(B) The pitch of a sound depends on the frequency of the vibration. Since the same key is struck,the frequency remains constant,and therefore the pitch does not change.
Loudness depends on the amplitude of the vibration. When the key is struck harder,the amplitude of the vibration increases,which results in a louder sound.

(C) $SONAR$ stands for $\text{Sound Navigation and Ranging}$.
It is a technique used to determine the direction, distance, and speed of underwater objects.
$SONAR$ works on the principle of reflection of sound waves.
It specifically uses ultrasonic waves (sound waves with frequencies higher than $20,000 \text{ Hz}$) because these waves can travel through water for long distances without being scattered significantly.

(D) Sound is a mechanical wave that propagates through a medium by the transfer of energy.
During the propagation of sound,the particles of the medium do not travel from one place to another; they only oscillate about their mean positions.
It is the disturbance (energy) that travels from one particle to the next through the medium.

(A) The loudness of a sound is directly proportional to the square of the amplitude of the vibration producing the sound. Therefore,when we change a feeble sound to a loud sound,we increase its amplitude.

(A) The wavelength $(\lambda)$ is defined as the distance between two consecutive crests (peaks) or two consecutive troughs.
In the given wave diagram, the distance $AE$ represents one complete wavelength, as it covers one full crest and one full trough.
A half wavelength $(\lambda/2)$ corresponds to the distance of either one crest or one trough.
Looking at the diagram, the distance $AB$ represents one crest, and the distance $DE$ represents one trough.
However, the question asks for half the wavelength. Since $AB$ is the length of one crest, it represents $\lambda/2$. Similarly, $DE$ also represents $\lambda/2$.
Given the options, $AB$ is a correct representation of half the wavelength.

(C) Earthquakes produce low-frequency vibrations known as infrasound before the main shock wave begins.
These infrasonic waves are below the human audible range (less than $20 \ Hz$).
Some animals are sensitive to these low-frequency vibrations and show strange changes in their behavior,which serves as an advanced warning of the earthquake.

(D) Infrasound refers to sound waves with frequencies below the audible range of humans,which is typically below $20\, Hz$. Rhinoceros communicate using infrasound at frequencies as low as $5\, Hz$.

(A) The frequency of a particular musical instrument should be in tune with that of other musical instruments to ensure harmony.
By adjusting the tension of the strings,the sitarist changes the frequency of vibration of the strings.
This process,known as tuning,ensures that the sitar produces notes that are in harmony with the other instruments in the orchestra,resulting in pleasant music.

(A) From the graph,the time period $T$ is the time taken for one complete oscillation.
Looking at the graph,one complete wave cycle spans from $t = 0.5 \, \mu s$ to $t = 2.5 \, \mu s$ (or similar interval),which gives $T = 2 \, \mu s = 2 \times 10^{-6} \, s$.
Frequency $f = 1/T = 1 / (2 \times 10^{-6} \, s) = 0.5 \times 10^6 \, Hz = 5 \times 10^5 \, Hz$.
Given velocity $v = 1500 \, m \, s^{-1}$.
Wavelength $\lambda = v / f = 1500 / (5 \times 10^5) = 300 \times 10^{-5} \, m = 3 \times 10^{-3} \, m$.

(A) Graph $(a)$ is likely to be the male voice.
Reason: The pitch of a sound is determined by its frequency. $A$ lower frequency corresponds to a lower pitch. In the given graph,wave $(a)$ has a lower frequency (fewer oscillations per unit time) compared to wave $(b)$. Since the male voice typically has a lower pitch than the female voice,graph $(a)$ represents the male voice.

(D) To hear an echo,the time interval between the original sound and the reflected sound must be at least $0.1 \,s$.
The minimum distance required for the reflected sound wave to be heard as an echo is calculated as:
Distance = Velocity of sound $\times$ Time interval
Distance = $344 \,m/s \times 0.1 \,s = 34.4 \,m$.
In this scenario,the girl is in the middle of a $12 \,m$ wide park. The sound travels from the road to the building $(6 \,m)$ and reflects back to the girl $(6 \,m)$.
Total distance travelled by the reflected sound = $6 \,m + 6 \,m = 12 \,m$.
Since $12 \,m < 34.4 \,m$,the reflected sound reaches the girl in less than $0.1 \,s$. Therefore,the girl cannot hear a distinct echo.

(A) The sound produced by humming bees is caused by the rapid vibration of their wings,which creates a frequency within the human audible range ($20 \,Hz$ to $20,000 \,Hz$).
In contrast,the vibrations of a typical pendulum are very slow,resulting in a frequency below $20 \,Hz$,which is classified as infrasonic sound.
Since the human ear cannot detect frequencies below $20 \,Hz$,we cannot hear the sound of the pendulum's vibrations.

(B) When an explosion occurs underwater,it creates a sudden pressure disturbance that travels through the medium.
Since water is a fluid,it can only support the propagation of pressure variations in the form of longitudinal waves.
Sound waves are mechanical waves that require a medium to travel,and in fluids like water,they propagate as longitudinal waves where the particles of the medium oscillate parallel to the direction of wave propagation.

(C) The speed of sound is given as $v = 340 \, m \, s^{-1}$.
The time interval between seeing the lightning and hearing the thunder is $t = 10 \, s$.
Since the speed of light is extremely high,the time taken for light to reach the observer is negligible.
Therefore,the distance $d$ of the thunder cloud can be calculated using the formula: $d = v \times t$.
Substituting the values: $d = 340 \, m \, s^{-1} \times 10 \, s = 3400 \, m$.
Thus,the approximate distance of the thunder cloud is $3400 \, m$.

(D) According to the laws of reflection,the angle of incidence is equal to the angle of reflection.
The incident sound wave makes an angle of $50^o$ with the reflecting surface (the wall).
The normal is perpendicular to the reflecting surface,so the angle between the normal and the surface is $90^o$.
Therefore,the angle of incidence $i = 90^o - 50^o = 40^o$.
Since the angle of incidence equals the angle of reflection $(i = r)$,the angle of reflection $r$ is also $40^o$.
In the figure,$x$ represents the angle of reflection with respect to the normal.
Thus,$x = 40^o$.

(B) The ceiling and the wall behind the stage in good conference halls or concert halls are made curved so that sound,after reflection from these surfaces,spreads evenly across the entire hall. This ensures that the sound reaches all members of the audience clearly and effectively without significant loss of intensity.

(N/A) The graphical representation of sound waves involves plotting amplitude on the $y$-axis against time on the $x$-axis.
$(i)$ For two sound waves with the same amplitude but different frequencies,the height of the peaks (amplitude) remains constant for both waves,but the number of oscillations per unit time (frequency) differs.
$(ii)$ For two sound waves with the same frequency but different amplitudes,the number of oscillations per unit time is identical,but the maximum displacement from the mean position (amplitude) varies.
$(iii)$ For two sound waves with different amplitudes and different wavelengths,both the height of the peaks and the distance between consecutive crests (wavelength) differ between the two waves.

(C) The relationship between the speed of sound $(v)$, frequency $(\nu)$, and wavelength $(\lambda)$ is given by the formula: $v = \nu \lambda$.
$(i)$ Given: $v = 340 \, m \, s^{-1}$, $\nu = 256 \, Hz$.
Using $v = \nu \lambda$, we get $340 = 256 \times \lambda$.
$\lambda = \frac{340}{256} = 1.328 \, m \approx 1.33 \, m$.
$(ii)$ Given: $v = 340 \, m \, s^{-1}$, $\lambda = 0.85 \, m$.
Using $v = \nu \lambda$, we get $340 = \nu \times 0.85$.
$\nu = \frac{340}{0.85} = 400 \, Hz$.

(N/A) The curve representing density or pressure variations with respect to distance for a sound wave is shown in the provided image.
$1$. Compression: These are the regions of high pressure and high density,represented by the crests of the wave.
$2$. Rarefaction: These are the regions of low pressure and low density,represented by the troughs of the wave.
$3$. Wavelength: It is defined as the distance between two consecutive compressions or two consecutive rarefactions.
$4$. Time-period: It is defined as the time taken by the wave to travel the distance between any two consecutive compressions or rarefactions from a fixed point.

(B) wave is a disturbance in a medium caused by the repeated periodic motion of particles about their mean position.
In this process,energy is transferred from one point to another through the medium.
Crucially,the particles of the medium only oscillate about their equilibrium positions and do not undergo actual net displacement from one location to another.

(N/A) The two types of mechanical waves are:
$(i)$ Transverse waves: In these waves,the particles of the medium vibrate in a direction perpendicular to the direction of wave propagation.
$(ii)$ Longitudinal waves: In these waves,the particles of the medium vibrate in a direction parallel to the direction of wave propagation.

(B) transverse wave is a type of wave in which the particles of the medium oscillate or vibrate in a direction perpendicular to the direction of the wave's propagation.
For example,waves produced on a stretched string or ripples on the surface of water are transverse waves.

(B) longitudinal wave is a type of mechanical wave in which the particles of the medium oscillate or vibrate back and forth in the same direction as the wave propagation.
This results in the formation of regions of high pressure called compressions and regions of low pressure called rarefactions.
Sound waves in air are the most common example of longitudinal waves.

(B) longitudinal wave is a type of wave in which the particles of the medium oscillate or vibrate back and forth in the same direction as the wave propagates.
This means the displacement of the medium is parallel to the direction of the wave's travel,resulting in regions of compressions and rarefactions.

(B) crest is defined as the point of maximum positive displacement in a transverse wave.
It represents the highest point or the peak of the wave above the mean equilibrium position.
In contrast,the point of maximum negative displacement is known as a trough.

(B) trough is the point of maximum negative displacement in a wave.
It represents the lowest point of a wave below the mean or equilibrium position.
In a transverse wave,the crests are the high points and the troughs are the low points.

(B) Sound is a form of mechanical energy that travels through a medium as a wave.
It is produced by the vibration of objects.
When these vibrations reach our ears,they produce a sensation of hearing,allowing us to perceive sound.

(N/A) Infrasonic vibrations,also known as infrasound,refer to sound waves that have a frequency below the lower limit of human audibility,which is $20 \,Hz$. These vibrations are typically produced by sources such as earthquakes,volcanic eruptions,and the movement of large animals like whales and elephants.

(B) The audible range for human beings is the range of frequencies that the human ear can detect. This range is typically from $20 \,Hz$ to $20,000 \,Hz$. Frequencies below $20 \,Hz$ are called infrasonic,and frequencies above $20,000 \,Hz$ are called ultrasonic.

(N/A) $(i)$ One complete to and fro motion of a body is called one vibration.
$(ii)$ The time taken by a vibrating body to complete one vibration is called its time period.
$(iii)$ The number of complete vibrations that the vibrating body makes in one second is called its frequency.
$(iv)$ The maximum displacement of a vibrating body from its mean position is called its amplitude.

(B) The human ear can hear sound waves in the audible frequency range,which is between $20 \, Hz$ and $20,000 \, Hz$ $(20 \, kHz)$.
$1$. $10 \, Hz$ is infrasonic (below $20 \, Hz$),so it cannot be heard.
$2$. $500 \, Hz$,$1500 \, Hz$,and $12000 \, Hz$ fall within the audible range $(20 \, Hz - 20,000 \, Hz)$.
$3$. $25000 \, Hz$ is ultrasonic (above $20,000 \, Hz$),so it cannot be heard.
Therefore,we can hear $500 \, Hz, 1500 \, Hz,$ and $12000 \, Hz$.

(C) Ultrasonic vibrations,or ultrasonics,refer to sound waves that have a frequency higher than the upper limit of human hearing,which is $20,000 \,Hz$ (or $20 \,kHz$).
These vibrations are beyond the audible range for humans.

(N/A) Animals that can hear ultrasonic vibrations (frequencies greater than $20,000 \ Hz$) include dogs,bats,and dolphins. These animals possess specialized auditory systems that allow them to perceive high-frequency sounds beyond the range of human hearing.

(N/A) Elastic waves are mechanical waves that propagate through a material medium due to the elasticity and inertia of the medium's particles.
These waves require a medium to travel and cannot propagate through a vacuum.
The two primary types of elastic waves are:
$1$. Longitudinal waves: Waves in which the particles of the medium oscillate parallel to the direction of wave propagation.
$2$. Transverse waves: Waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation.

(N/A) An echo is a repetition of sound caused by the reflection of sound waves from a distant,rigid obstacle,such as a wall,cliff,or mountain. For an echo to be heard distinctly by the human ear,the reflected sound must reach the listener at least $0.1 \ s$ after the original sound.

(N/A) Echoes are used in the following ways:
$1$. $SONAR$ ($Sound$ $Navigation$ $And$ $Ranging$) systems use echoes to determine the depth of the sea and to locate underwater objects like submarines,shipwrecks,or icebergs.
$2$. Medical ultrasound imaging uses echoes of high-frequency sound waves to create images of internal organs,such as monitoring the development of a fetus during pregnancy.

(A) The speed of light in air is approximately $3 \times 10^{8} \text{ m s}^{-1}$,whereas the speed of sound in air is approximately $340 \text{ m s}^{-1}$ at room temperature. Since the speed of light is significantly higher than the speed of sound,the flash of lightning is observed by our eyes much earlier than the sound of thunder reaches our ears.

(B) Reverberation is defined as the repeated reflection of sound waves from the walls,ceiling,and floor of an enclosed space.
This phenomenon causes the sound to persist for a certain duration even after the source of the sound has stopped emitting it.
It is commonly observed in large halls or auditoriums with hard,reflective surfaces.

(B) Frequency is defined as the number of oscillations or waves produced per unit of time.
Given that $20$ waves are produced in $1$ second.
Therefore,the frequency $f = \frac{\text{Number of waves}}{\text{Time}} = \frac{20}{1} = 20\, Hz$.

(B) $SONAR$ stands for $\text{Sound Navigation And Ranging}$.
It is a technique that uses ultrasonic sound waves to determine the depth of the sea and to locate underwater objects such as reefs,submarines,and schools of fish.
The system works by emitting sound pulses and measuring the time taken for the echo to return after reflecting off an object.

(N/A) Sound waves with a frequency greater than $20,000 \,Hz$ are known as ultrasonic waves or ultrasound.
These waves are widely used in medical diagnostics,such as in ultrasonography (ultrasound scans) to visualize internal organs of the human body.

(N/A) To hear an echo,the minimum distance between the source of sound and the reflecting surface (obstacle) must be approximately $17 \,m$. This is because the human brain retains sound for $0.1 \,s$ (persistence of hearing). Since the speed of sound in air is about $344 \,m/s$,the total distance traveled by sound in $0.1 \,s$ is $34.4 \,m$. Therefore,the distance to the obstacle must be at least $34.4 / 2 = 17.2 \,m$. In small rooms,this distance is not maintained,so the reflected sound reaches the ear before the $0.1 \,s$ interval,causing the original sound and the echo to overlap,making the echo inaudible.

(A) An echo is the repetition of sound caused by the reflection of sound waves from a hard surface.
For an echo to be heard distinctly,the time interval between the original sound and the reflected sound must be at least $0.1 \, s$.
Since the speed of sound in air is approximately $344 \, m/s$,the total distance traveled by the sound in $0.1 \, s$ is $34.4 \, m$.
Therefore,the minimum distance between the source of sound and the reflecting surface must be half of this,which is $17.2 \, m$.
In long galleries and big halls,the dimensions are large enough to provide this minimum distance,allowing the reflected sound to reach the listener after the required time interval,thus creating an echo.

(N/A) Sound requires a material medium for its propagation. The moon has no atmosphere,meaning it is a vacuum. Since sound waves cannot travel through a vacuum,astronauts cannot hear each other on the moon.

(N/A) The speed of light in air is approximately $3 \times 10^8 \text{ m/s}$, whereas the speed of sound in air is approximately $343 \text{ m/s}$ at room temperature.
Because the velocity of light is significantly greater than the velocity of sound, the light from the lightning flash reaches the observer almost instantaneously.
In contrast, the sound waves produced by the thunder travel much slower, resulting in a noticeable time delay between seeing the flash and hearing the thunder.

(B) Bats possess the ability to navigate and hunt in the dark using a biological sonar system known as echolocation.
When bats fly,they emit high-frequency sound waves called ultrasonic waves.
These waves travel through the air and strike nearby objects,reflecting back to the bat's ears.
By analyzing the time delay,intensity,and frequency shift of the returning echoes,the bat can accurately determine the distance,direction,size,and texture of the objects in its path.
This highly sensitive auditory system allows them to maneuver and hunt effectively without the need for vision.

(A) $(i)$ Light waves are electromagnetic waves,which are transverse in nature because their oscillations occur perpendicular to the direction of wave propagation.
$(ii)$ Waves in water (surface waves) are transverse waves because the water particles move up and down perpendicular to the direction of the wave travel.
$(iii)$ Sound waves are longitudinal waves because the particles of the medium vibrate back and forth parallel to the direction of wave propagation.

(N/A) The speed of sound in a medium depends on the following three factors:
$(i)$ Density of the medium: The speed of sound is inversely proportional to the square root of the density of the medium.
$(ii)$ Elasticity of the medium: The speed of sound is directly proportional to the square root of the elasticity of the medium. Sound travels faster in more elastic media.
$(iii)$ Temperature of the medium: The speed of sound in a gas is directly proportional to the square root of its absolute temperature (measured in Kelvin).

(N/A) $(i)$ The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of a wave.
$(ii)$ The relationship between speed of sound $(V)$,wavelength $(\lambda)$,and frequency $(\nu)$ is given by the formula: $V = \lambda \times \nu$.