Mix Example - SOUND Questions in English

(N/A) The persistence of sound in an auditorium as a result of repeated reflections of sound is called reverberation. Two ways to reduce reverberation are: $(i)$ Covering the walls and ceiling of the auditorium with sound-absorbing materials like compressed fibreboard,rough plaster,or draperies. $(ii)$ Using sound-absorbing materials for seats and carpets.
$(b)$ $A$ sound of a single frequency is called a tone. $A$ sound which is produced due to a mixture of several frequencies is called a note.
$(c)$ Ultrasound can be used to detect cracks and flaws in metal blocks. Metallic components are generally used in the construction of big structures like buildings,bridges,machines,and scientific equipment. The cracks or holes inside the metal blocks,which are invisible from the outside,reduce the strength of the structure. Ultrasonic waves are allowed to pass through the metal block,and detectors are used to detect the transmitted waves. If there is even a small defect,the ultrasound gets reflected back,indicating the presence of the flaw or defect,as shown in the figure.

(N/A) The angle between the tube and the reflecting surface is $50^{\circ}$. The normal is perpendicular to the surface,so the angle of reflection $r$ is $90^{\circ} - 50^{\circ} = 40^{\circ}$. According to the laws of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$. Therefore,$\angle x = \angle r = 40^{\circ}$ and $\angle y = \angle i = 40^{\circ}$.
$(b)$ The phenomenon observed is the reflection of sound.
$(c)$ The laws of reflection of sound are:
$(i)$ The incident sound wave,the reflected sound wave,and the normal at the point of incidence all lie in the same plane.
$(ii)$ The angle of incidence is always equal to the angle of reflection $(\angle i = \angle r)$.

(A-D) Given: $\lambda = 420.5\, m$,$V = 3 \times 10^{8}\, m s^{-1}$.
Using the wave equation $V = \nu \lambda$,the frequency $\nu$ is given by $\nu = V / \lambda$.
Substituting the values: $\nu = (3 \times 10^{8}) / 420.5 \approx 7.13 \times 10^{5}\, Hz$.
$(b)$ $(i)$ In a transverse wave,the particles of the medium oscillate perpendicular to the direction of wave propagation.
$(ii)$ In a longitudinal wave,the particles of the medium oscillate parallel to the direction of wave propagation.

(N/A) Given: Speed of sound in air $V_{a} = 340 \, m s^{-1}$,frequency $v = 20 \, Hz$.
Using the formula $V = v \lambda$,the wavelength in air is $\lambda_{a} = \frac{V_{a}}{v} = \frac{340}{20} = 17 \, m$.
Given: Speed of sound in water $V_{w} = 1480 \, m s^{-1}$.
The frequency $v$ remains constant when the source is moved to a different medium.
Therefore,the wavelength in water is $\lambda_{w} = \frac{V_{w}}{v} = \frac{1480}{20} = 74 \, m$.

(N/A) Given: Time period $T = 5 \times 10^{-3} \text{ s}$,Velocity $V = 1450 \text{ m s}^{-1}$.
$1$. Frequency $(v)$: The frequency is the reciprocal of the time period.
$v = \frac{1}{T} = \frac{1}{5 \times 10^{-3}} = \frac{1000}{5} = 200 \text{ Hz}$.
$2$. Wavelength $(\lambda)$: The wavelength is calculated using the relation $V = v \times \lambda$.
$\lambda = \frac{V}{v} = \frac{1450}{200} = 7.25 \text{ m}$.
Thus,the frequency is $200 \text{ Hz}$ and the wavelength is $7.25 \text{ m}$.

(A) Given: Frequency $(v) = 400 \, Hz$, Speed of sound $(V) = 340 \, m/s$, Number of vibrations $(n) = 16$.
First, calculate the wavelength $(\lambda)$ of the sound wave using the formula: $\lambda = V / v$.
$\lambda = 340 / 400 = 0.85 \, m$.
The distance traveled by sound in one vibration is equal to its wavelength $(\lambda)$.
Therefore, the total distance traveled in $16$ vibrations is $d = n \times \lambda$.
$d = 16 \times 0.85 = 13.6 \, m$.
Thus, the sound travels $13.6 \, m$ when the tuning fork makes $16$ vibrations.

(A) The distance of the wall is $d = 33\, m$.
The velocity of sound is $v = 330\, m s^{-1}$.
The time taken for the sound to travel to the wall and back is given by the formula $t = \frac{2d}{v}$.
Substituting the values,$t = \frac{2 \times 33}{330} = \frac{66}{330} = 0.2\, s$.
For an echo to be heard by the human ear,the time interval between the original sound and the reflected sound must be at least $0.1\, s$.
Since the calculated time $0.2\, s$ is greater than $0.1\, s$,the echo will be heard.

$300 \, m \, s^{-1}$ In a transverse wave, the distance between a crest and the consecutive trough is equal to half of the wavelength $(\lambda/2)$.
Given, $\lambda/2 = 15 \, cm$.
Therefore, the wavelength $\lambda = 2 \times 15 = 30 \, cm = 0.3 \, m$.
The frequency $f = 1000 \, Hz$.
The velocity of the wave $V$ is given by the formula $V = f \times \lambda$.
Substituting the values, $V = 1000 \times 0.3 = 300 \, m \, s^{-1}$.

(N/A) The velocity $(v)$ of the sound wave is calculated by dividing the total distance traveled by the time taken:
$v = \frac{\text{Distance}}{\text{Time}} = \frac{840 \ m}{2.5 \ s} = 336 \ m/s$.
The frequency $(f)$ of the sound wave is calculated using the formula $v = f \times \lambda$,where $\lambda$ is the wavelength.
Given $\lambda = 70 \ cm = 0.7 \ m$ and $v = 336 \ m/s$:
$f = \frac{v}{\lambda} = \frac{336 \ m/s}{0.7 \ m} = 480 \ Hz$.

(B) The distance between the person and the building is $d = 167 \ m$.
The sound travels to the building and reflects back to the person,covering a total distance of $2d = 2 \times 167 \ m = 334 \ m$.
The speed of sound is given as $V = 334 \ m \ s^{-1}$.
The time taken to hear the echo is calculated using the formula $t = \frac{\text{Total Distance}}{\text{Speed}} = \frac{2d}{V}$.
Substituting the values: $t = \frac{334 \ m}{334 \ m \ s^{-1}} = 1.0 \ s$.
Therefore,the person hears the echo after $1.0 \ s$.

(A) The distance of the person from the building is $d = 132.8 \, m$.
The total distance traveled by the sound to the building and back to the person is $2d = 2 \times 132.8 = 265.6 \, m$.
The time taken to hear the echo is $t = 0.8 \, s$.
The speed of sound $V$ is given by the formula $V = \frac{\text{Total distance}}{\text{Time}} = \frac{2d}{t}$.
Substituting the values: $V = \frac{265.6}{0.8} = 332 \, m \, s^{-1}$.

(A) The speed of sound $V = 310 \, m s^{-1}$.
The time taken for the echo to be heard is $t = 1.5 \, s$.
Since the sound travels to the cliff and back to the person,the total distance covered is $2d$,where $d$ is the distance of the person from the cliff.
The formula for speed is $V = \frac{2d}{t}$.
Rearranging for distance: $2d = V \times t$.
Substituting the values: $2d = 310 \times 1.5 = 465 \, m$.
Therefore,the distance $d = \frac{465}{2} = 232.5 \, m$.

(N/A) Let the speed of sound be $v = 332 \, m/s$.
The time taken for the first echo is $t_1 = 0.4 \, s$. The distance of the person from the nearer building $(d_1)$ is given by:
$d_1 = \frac{v \times t_1}{2} = \frac{332 \times 0.4}{2} = 66.4 \, m$.
The time taken for the second echo is $t_2 = 2.0 \, s$. The distance of the person from the farther building $(d_2)$ is given by:
$d_2 = \frac{v \times t_2}{2} = \frac{332 \times 2.0}{2} = 332 \, m$.
The total distance between the two buildings is the sum of the distances from the person to each building:
$D = d_1 + d_2 = 66.4 + 332 = 398.4 \, m$.

(N/A) Given: Time period $T = 0.01 \, s$,Speed $V = 15 \, m s^{-1}$.
Using the relation $V = \nu \lambda$ and $\nu = 1/T$,we have $V = \lambda / T$.
Therefore,the wavelength $\lambda = V \times T = 15 \times 0.01 = 0.15 \, m$.
The distance between an adjacent crest and a trough is equal to half of the wavelength $(\lambda / 2)$.
Distance $= 0.15 / 2 = 0.075 \, m$.

(A) Given: Speed of sound $V = 1.7 \, km \, s^{-1} = 1700 \, m \, s^{-1}$.
Frequency $\nu = 4.2 \, MHz = 4.2 \times 10^{6} \, Hz$.
Wavelength $\lambda = ?$.
Using the wave equation $V = \nu \lambda$,we can calculate the wavelength as $\lambda = V / \nu$.
Substituting the values: $\lambda = 1700 / (4.2 \times 10^{6}) \, m$.
$\lambda \approx 4.047 \times 10^{-4} \, m$,which is approximately $4 \times 10^{-4} \, m$.

(A) Given: Time taken for the echo to return,$t = 6 \, s$. Speed of sound,$V = 342 \, m s^{-1}$.
An echo is the sound heard after reflection from a distant surface. The sound travels to the reflecting surface and back to the source,covering a total distance of $2d$ in time $t$.
The formula for distance is: $d = \frac{V \times t}{2}$.
Substituting the values: $d = \frac{342 \times 6}{2} = 342 \times 3 = 1026 \, m$.
Therefore,the distance of the reflecting surface from the source is $1026 \, m$.

(N/A) The frequency $(v)$ is defined as the number of waves passing through a point per unit time.
Given that $20$ waves pass in $2$ seconds,the frequency is calculated as:
$v = \frac{20 \text{ waves}}{2 \text{ seconds}} = 10 \text{ Hz}$.
$(b)$ The distance between a crest and the adjacent trough is equal to half of the wavelength $(\frac{\lambda}{2})$.
Given that the distance is $1.5 \, m$,we have:
$\frac{\lambda}{2} = 1.5 \, m$.
Therefore,the wavelength $\lambda = 1.5 \times 2 = 3 \, m$.

(B) The diver will hear the sound first. This is because the speed of sound in water (a liquid medium) is significantly higher than the speed of sound in air (a gaseous medium). Since the distance is the same for both,the sound travels faster through the water to reach the diver.
$(b)$ Given:
Frequency $(v) = 220 \ Hz$
Speed $(V) = 440 \ m \ s^{-1}$
Formula: $V = v \times \lambda$
Therefore,wavelength $(\lambda) = V / v$
$\lambda = 440 / 220 = 2 \ m$
The wavelength of the sound wave is $2 \ m$.

(N/A) Number of waves produced in $3 \, s = 1500$.
Number of waves produced in $1 \, s = 1500 / 3 = 500$.
Therefore,frequency,$\nu = 500 \, Hz$.
$(b)$ Wavelength is defined as the distance between two consecutive compressions or two consecutive rarefactions. The distance between a compression and an adjacent rarefaction is half the wavelength $(\lambda / 2)$.
Given,$\lambda / 2 = 68 \, cm$.
Therefore,wavelength,$\lambda = 68 \times 2 = 136 \, cm = 1.36 \, m$.
$(c)$ Velocity,$V = \nu \times \lambda$.
$V = 500 \, Hz \times 1.36 \, m = 680 \, m/s$.

(B) Given: Speed of sound $(V) = 399\, m s^{-1}$,Wavelength $(\lambda) = 1.5\, cm = 0.015\, m$.
We know the relationship between speed,frequency $(v)$,and wavelength $(\lambda)$ is given by the formula: $V = v \times \lambda$.
Rearranging for frequency: $v = V / \lambda$.
Substituting the values: $v = 399 / 0.015 = 26600\, Hz$.
The human audible range is between $20\, Hz$ and $20,000\, Hz$.
Since the calculated frequency of $26600\, Hz$ is greater than $20,000\, Hz$,it is ultrasonic and therefore not audible to the human ear.

(A) Given: Speed of sound in water $(V)$ = $1530 \, m s^{-1}$.
Total time taken for the echo to return $(t)$ = $80 \, s$.
Let the distance of the obstacle from the submarine be $d$.
The sound travels to the obstacle and back to the submarine,covering a total distance of $2d$.
Using the formula: $2d = V \times t$.
Substituting the values: $2d = 1530 \times 80$.
$2d = 122400 \, m$.
Therefore,$d = 122400 / 2 = 61200 \, m$.

(A) Given:
Frequency $(f)$ = $400 \, Hz$
Speed of sound $(V)$ = $340 \, m s^{-1}$
Number of vibrations $(n)$ = $16$
The time period $(T)$ for one vibration is $T = 1/f = 1/400 \, s$.
Total time $(t)$ taken for $16$ vibrations is $t = n \times T = 16 \times (1/400) = 16/400 \, s$.
Distance $(d)$ traveled by sound is calculated using the formula $d = V \times t$.
Substituting the values: $d = 340 \times (16/400) = 340 \times 0.04 = 13.6 \, m$.
Therefore,the sound travels $13.6 \, m$.

(A) Given: Frequency $(v)$ = $4.2 \times 10^{6} \text{ Hz}$,Speed of sound $(V)$ = $1700 \text{ m s}^{-1}$.
We know the relationship between speed,frequency,and wavelength is given by the formula: $V = v \lambda$.
To find the wavelength $(\lambda)$,we rearrange the formula: $\lambda = V / v$.
Substituting the given values: $\lambda = 1700 / (4.2 \times 10^{6})$.
$\lambda = 404.76 \times 10^{-6} \text{ m} \approx 4.04 \times 10^{-4} \text{ m}$.

(N/A) This happens when the frequency of vibration produced by the engine of the motorbike becomes equal to the natural frequency of the rear view mirror.
$(b)$ The name of this phenomenon is resonance.
$(c)$ The speed of the motorbike should be changed,which will alter the frequency of vibration of the engine,thereby stopping the resonance condition.

(N/A) $(i)$ The phenomenon is called resonance.
$(ii)$ The sound becomes loudest because the natural frequency of the air column becomes equal to the frequency of the vibrating tuning fork.
$(iii)$ The frequency of the loud sound is equal to the frequency of the tuning fork.
$(iv)$ The unit for measuring loudness is the decibel $(dB)$.

(NO) The quality of the two notes need not be the same.
The pitch of a note is determined by its frequency.
The quality (or timbre) of a sound depends upon the number,distribution,and relative intensity of the different harmonics and overtones produced by the instrument.
Even if two guitars are identical,the way they are played,the material of the strings,and the resonance of the wooden body can vary,leading to differences in the overtones produced,thus resulting in different sound qualities.

(N/A) $(i)$ Yes,the above action produces an audible sound because the surface area of the table is much larger than the tuning fork,which helps in vibrating a larger volume of air.
$(ii)$ Yes,the above action causes the table top to set into vibrations.
$(iii)$ These vibrations are known as forced vibrations.
$(iv)$ Resonance occurs when the frequency of the tuning fork is equal to the natural frequency of oscillation of the table top.

(N/A) The 'quality' or 'timbre' of a sound is determined by the presence and relative intensity of overtones or harmonics.
Even if two strings produce notes of the same fundamental frequency (pitch) and amplitude (loudness),they may vibrate in different modes or have different harmonic structures depending on the material,tension,and point of plucking.
These additional harmonics superimpose on the fundamental frequency,creating a unique waveform for each string.
Therefore,the overall effect of these harmonics results in a different 'quality' of sound for each string.

(N/A) $(i)$ Yes,an audible sound is produced because the large surface area of the table forces more air molecules to vibrate,increasing the intensity of the sound.
$(ii)$ Yes,the table top is set into vibrations by the stem of the tuning fork.
$(iii)$ These vibrations are known as 'forced vibrations' because the table is forced to vibrate at the frequency of the tuning fork.
$(iv)$ Resonance occurs when the frequency of the vibrating tuning fork exactly matches the natural frequency of the table top,resulting in a significantly louder sound.

(N/A) $(i)$ In mode $(A)$,the wire vibrates in one loop,so the length $L = \lambda/2$,which means $\lambda = 2L = 2 \times 0.5\, m = 1\, m$. In mode $(B)$,the wire vibrates in two loops,so the length $L = \lambda'$. Thus,$\lambda' = L = 0.5\, m$. Therefore,the wavelength in mode $(B)$ is $0.5\, m$.
$(ii)$ The note produced in mode $(A)$ is louder. This is because the amplitude (maximum displacement from the mean position) of the wave in mode $(A)$ is greater than that in mode $(B)$. Since loudness is directly proportional to the square of the amplitude,the sound in mode $(A)$ is louder.
$(iii)$ The pitch of the note produced in mode $(B)$ is higher. This is because pitch is directly related to the frequency of the sound wave. As seen from the diagram,mode $(B)$ has two loops,meaning it vibrates at a higher frequency compared to mode $(A)$,which has only one loop. Higher frequency results in a higher pitch.

(N/A) The characteristics of a wave motion are:
$(i)$ It is a periodic disturbance that travels through a medium.
$(ii)$ Energy and momentum are transferred from one point to another without the actual transfer of matter.

(B) Loudness is directly proportional to the square of the amplitude of the wave $(L \propto A^2)$.
When the amplitude $(A)$ is doubled $(2A)$, the new loudness becomes proportional to $(2A)^2 = 4A^2$.
Therefore, the loudness becomes four times the original value.

(N/A) The $frequency$ of a musical sound determines its $pitch$. $A$ higher frequency results in a $sharper$ and $shriller$ sound.
The $amplitude$ of a musical sound determines its $loudness$ or $intensity$. $A$ larger amplitude results in a $louder$ or $more$ $intense$ sound.

(N/A) Even though the two musical notes have the same pitch (frequency) and loudness (amplitude),they sound different because of their different 'quality' or 'timbre'.
This difference in quality arises because the number and nature of the harmonics and overtones produced by the violin and the piano are different.
Each musical instrument produces a unique combination of harmonics and overtones along with the fundamental frequency,which results in the distinct wave patterns observed for each instrument.

(N/A) The subjective property of sound waves,which is related to its frequency,is known as its $pitch$.
As the frequency of sound increases,the $pitch$ also increases.
The subjective property of light,which is related to its wavelength,is known as the $colour$ of light.
Different wavelengths of light correspond to different perceived $colours$ in the visible spectrum.

(NO) The quality of the two notes need not be the same.
This is because while the pitch of a note is determined solely by its frequency,the quality (or timbre) is determined by its harmonic or overtone content.
It is the number,distribution,and relative intensity of the different harmonics and overtones that determine the quality of a sound.
Even if two guitars are identical,the way they are played or slight variations in the construction and material of the strings can lead to differences in the harmonic content,meaning the quality of the notes produced may differ.

(N/A) $(i)$ Natural vibration: The oscillations of a simple pendulum about its mean position in the absence of any external force.
$(ii)$ Forced vibration: $A$ sonometer wire,kept under tension,vibrating under the influence of an external vibrating tuning fork.
$(iii)$ Resonance: When the frequency of an external periodic force matches the natural frequency of a system,such as a sonometer wire vibrating with maximum amplitude when the tuning fork frequency matches its own natural frequency.

(N/A) Echoes are used in $SONAR$ (Sound Navigation and Ranging) and $RADAR$ systems to detect the presence and estimate the distance of underwater objects or flying objects,respectively.

(N/A) When a stretched string is set into vibration,it pushes and pulls the surrounding air molecules,creating a series of compressions and rarefactions. These vibrations travel through the air as sound waves. When these waves reach our ears,they cause the eardrum to vibrate,which the brain interprets as an audible sound.

(NO) No,the sound will not be audible on the surface of the moon.
Sound is a mechanical wave that requires a material medium (like air,water,or solids) to propagate.
The moon has no atmosphere,meaning it is a vacuum.
Since there is no medium for the sound waves to travel through,the vibrations cannot reach our ears,and therefore,no sound will be heard.

(A) $(i)$ When the frequency of a sound wave is increased,the pitch of the sound increases. Pitch is the characteristic of sound that determines the shrillness or gravity of a note.
(ii) When the amplitude of a sound wave is increased,the loudness of the sound increases. Loudness is a measure of the sound energy reaching the ear per unit time.

(N/A) The phenomenon is known as Echo.
An echo is the repetition of sound caused by the reflection of sound waves from a distant surface.
For an echo to be heard,the reflected sound must reach the observer at least $0.1 \, s$ after the original sound,as the human ear has a persistence of hearing of $0.1 \, s$.
The total distance traveled by the sound is $2d$,where $d$ is the distance to the reflecting surface.
Given the speed of sound in air is approximately $340 \, m/s$,the minimum distance required is calculated as $d = (v \times t) / 2 = (340 \times 0.1) / 2 = 17 \, m$.
Since the building is at a distance of $18 \, m$,which is greater than $17 \, m$,the reflected sound reaches the observer after $0.1 \, s$,allowing the echo to be heard clearly.

(C) wave is a disturbance that travels through a medium or vacuum,transferring energy from one point to another without the net transport of matter (mass).
While the particles of the medium oscillate about their mean positions,they do not travel along with the wave.
Therefore,energy is the physical quantity that is transferred by waves from one place to another.

(D) Sound waves are mechanical waves that require a medium for propagation.
In a longitudinal wave,the particles of the medium oscillate back and forth in the same direction as the wave propagation.
Since sound waves travel by creating compressions and rarefactions in the medium,they are classified as longitudinal waves.

(A) slinky is a long,flexible spring.
When we push and pull the slinky along its length,we produce longitudinal waves (compressions and rarefactions).
When we move the slinky up and down or side to side perpendicular to its length,we produce transverse waves (crests and troughs).
Therefore,both types of waves can be produced on a slinky.

(B) In a transverse wave,a crest represents the point of maximum positive displacement from the equilibrium position.
Consecutive crests are the two nearest points of maximum displacement in the same direction.
The distance between two consecutive crests or two consecutive troughs is defined as one wavelength (denoted by the Greek letter $\lambda$).

(C) The $SI$ unit of frequency is $Hertz$ $(Hz)$.
One $Hertz$ is defined as one cycle per second.
Frequency is defined as the number of oscillations or cycles completed by a wave in one second.
Therefore,$Hertz$ is the unit of frequency.

(D) Sound travels through a medium as a series of compressions and rarefactions.
Compressions are regions of high pressure and high density.
Rarefactions are regions of low pressure and low density.
Since the question specifies that the density of the medium is at its minimum,this point corresponds to a rarefaction.

(A) The audible range of sound for a healthy human ear is typically between $20 \, Hz$ and $20,000 \, Hz$ (or $20 \, kHz$).
Frequencies below $20 \, Hz$ are called infrasonic,and frequencies above $20,000 \, Hz$ are called ultrasonic.
Therefore,the correct range is $20 \, Hz$ to $20 \, kHz$.

(B) Ultrasonic waves are high-frequency sound waves $(> 20,000 \ Hz)$.
$SONAR$ (Sound Navigation and Ranging) uses ultrasonic waves to detect objects underwater.
Sonography (or ultrasound scanning) uses ultrasonic waves to create images of internal body structures.
$CUSA$ (Cavitron Ultrasonic Surgical Aspirator) uses ultrasonic energy to fragment and remove tissues during surgery.
Radio waves are a type of electromagnetic radiation,not sound waves,and do not involve ultrasonic technology. Therefore,ultrasonic waves are not used in radio waves.