SOUND-Exercise
Physics - Exercise
Q1. What is sound and how is it produced?
Solution:
Sound is a form of energy that travels in the form of vibrations through a medium such as air, water, or
solids. It produces the sensation of hearing when these vibrations reach our ears.
Sound is produced by vibrating objects. When an object vibrates, it causes the surrounding particles of the
medium to vibrate as well. These vibrating particles transfer energy to neighboring particles, creating a
series of compressions (regions of high pressure) and rarefactions (regions of low pressure). This
continuous disturbance travels outward as a sound wave.
For example, when you pluck a guitar string, the string vibrates rapidly. These vibrations set the nearby
air particles into motion, and the sound waves travel to your ear — allowing you to hear the note.
Q2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Solution:
When a source of sound, such as vibrating object, moves forward, it pushes and compresses the air in front of itto create a region of high pressure, called compression. As it moves backward, the air in front expands, leaving a region of low pressure, callled rarefaction. In this way, a series of compression and rarefaction is produced in the air, forming a sound wave.
Q3. Why is sound wave called a longitudinal wave?
Solution:
Sound waves are called longitudinal waves because the vibration of particles in the medium takes place in
the same direction as the direction of wave propagation.
When a sound is produced, the particles of the medium (like air) move back and forth along the line of
travel of the sound. This motion creates compressions (where particles are close together) and rarefactions
(where particles are farther apart).
Since these vibrations of particles are parallel to the direction in which the sound wave travels, the sound
wave is known as a longitudinal wave.
Q4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Solution:
The characteristic of sound that helps us identify our friend’s voice is called quality or timbre.
Every person’s voice produces sound waves with a unique pattern due to differences in their vocal cords,
throat shape, and mouth cavity. This gives their voice a distinct quality, even when the loudness and pitch
are the same as others.
Thus, in a dark room where we cannot see anyone, we can still recognise our friend’s voice because of its
unique timbre or quality of sound.
Q5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Solution:
Although both lightning (flash) and thunder (sound) are produced at the same time during a thunderstorm, we
see the flash first and hear the thunder after a few seconds because light travels much faster than
sound.
The speed of light in air is about \(3\times 10^8\,m/s\), while the speed of sound is only
about\(340\,m/s.\)
Because light travels almost instantly to our eyes, we see the flash right away, but the sound waves take
longer to reach our ears, causing the thunder to be heard later.
Q6. 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 \(344 \,m s^{–1}\).
Solution:
Given:
$$ \text{Lower frequency: } \nu_{l} = 20\,\mathrm{Hz} $$ $$ \text{Higher frequency: } \nu_{h} = 20\times10^3\,\mathrm{Hz} $$ $$ \text{Speed of sound: } v = 344\,\mathrm{m/s} $$Formula:
$$ v = \lambda \nu $$ \[ \implies \boxed{\lambda = \frac{v}{\nu}} \]Using the given formula, the wavelength for each frequency can be calculated as follows:
For the lower frequency:
$$ \begin{aligned} \lambda_l &= \frac{v}{\nu_l} \\ &= \frac{344}{20} \\ &= 17.2\,\mathrm{m} \end{aligned} $$For the higher frequency:
$$ \begin{aligned} \lambda_h &= \frac{v}{\nu_h} \\ &= \frac{344}{20\times10^3} \\ &= 0.0172\,\mathrm{m} \end{aligned} $$Final Answer:
$$ \boxed{\lambda_l = 17.2\,\mathrm{m}} \quad \text{and} \quad \boxed{\lambda_h = 0.0172\,\mathrm{m}} $$Hence, the wavelengths corresponding to the hearing range (20 Hz to 20 kHz) lie between 17.2 m and 0.0172 m.
Q7. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Solution:
Given:
Let the length of the aluminium rod be \( L \).
Speed of sound in air \( v_{\text{air}} = 344\,\mathrm{m/s} \)
Speed of sound in aluminium \( v_{\text{Al}} = 6420\,\mathrm{m/s} \)
To Find: Ratio of times taken by sound in air and in aluminium, i.e. \(\dfrac{t_{\text{air}}}{t_{\text{Al}}}\)
Formula:
$$ t = \frac{L}{v} $$So,
$$ \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{\left(\frac{L}{v_{air}}\right)}{\left(\frac{L}{v_{Al}}\right)} $$ \[ \implies \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{v_{\text{Al}}}{v_{\text{air}}} \]Substituting the values:
$$ \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{6420}{344} $$ \[ \boxed{\frac{t_{\text{air}}}{t_{\text{Al}}} \approx 18.66} \] The sound takes about 18.7 times longer to reach through air than through aluminium. $$ \boxed{\text{Ratio of times: } t_{\text{air}} : t_{\text{Al}} = 18.7 : 1} $$Q8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute
Solution:
Given:
Frequency of sound, \( f = 100\,\mathrm{Hz} \)
To Find:
Number of vibrations in one minute.
Solution:
Let the number of vibrations per minute be \( n \).
We know that frequency is the number of vibrations per second.
For one minute, \( t = 60\,\mathrm{s} \)
$$ \begin{aligned} 100 &= \frac{n}{60} \\ \implies n &= 100 \times 60 \\ \implies n &= 6000 \end{aligned} $$ $$ \boxed{n = 6000\,\text{vibrations per minute}} $$Hence, the source of sound vibrates 6000 times in one minute.
Q9. Does sound follow the same laws of reflection as light does? Explain.
Solution:
Yes, sound follows the same laws of reflection as light.
When sound waves strike a hard and smooth surface, they get reflected just like light waves do. The behavior
of
sound during reflection obeys two fundamental laws
- The angle of incidence is equal to the angle of reflection. \[\angle i=\angle r\]
- The incident wave, the reflected wave, and the normal at the point of incidence all lie in the same plane.
Thus, the reflection of sound is similar to the reflection of light, except that sound requires a material medium and a large, smooth surface for clear reflection.
Q10. 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 remain the same. Do you hear an echo sound on a hotter day?
Solution:
Yes, you will hear the echo sooner on a hotter day.
This is because the speed of sound increases with temperature. On a hot day, the air molecules move faster
and
transfer sound energy more quickly from one particle to another. As a result, the sound waves travel faster
through the air and return to the listener in less time after reflection.
Although the distance between the sound source and the reflecting surface remains the same, the time taken
for the echo to return decreases due to the higher speed of sound at higher temperatures.
The approximate relationship of sound speed with temperature in air is: \[v=331+0.6T\] where \(T\) is temperature in °C.
Q11. Give two practical applications of the reflection of sound waves.
Solution:
The reflection of sound is used in several practical applications. Two common examples are:
-
SONAR (Sound Navigation and Ranging):
Ships and submarines use SONAR to detect the depth of the sea or locate underwater objects. A sound pulse is sent into the water, and the time taken for the echo to return helps calculate the distance of the object. \[\text{Distance}=2\times t\] where \(v\) is the speed of sound in water and \(t\) is the time for the echo to return. -
Design of Auditoriums and Concert Halls:
In large halls and theaters, walls and ceilings are designed with curved or angled surfaces so that reflected sound waves reach all parts of the hall clearly. This ensures that everyone hears the speaker or performer equally well, reducing echo and improving sound quality.
Q12. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s–2 and speed of sound = 340 m s–1
Solution:
Let the time taken by the stone to fall to the water surface be \( t_1 \), and the time taken by the sound to travel back up be \( t_2 \).
Step 1: Time taken by the stone to fall
Using the equation of motion:
$$ s = \frac{1}{2} g t_1^2 $$ \[ 500 = \frac{1}{2} \times 10 \times t_1^2 \] \[ \implies t_1^2 = 100 \] \[ \boxed{t_1 = 10\,\mathrm{s}} \]Step 2: Time taken by the sound to travel up
$$ t_2 = \frac{\text{distance}}{\text{speed}} = \frac{500}{340} $$ \[ \boxed{t_2 = 1.47\,\mathrm{s}}\]Step 3: Total time after which the splash is heard
$$ t = t_1 + t_2 $$ \[ t = 10 + 1.47 = 11.47\,\mathrm{s} \]Final Answer:
$$ \boxed{t = 11.47\,\mathrm{seconds}} $$Hence, the splash is heard approximately 11.47 seconds after the stone is dropped.
Q13. 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?
Solution:
Given:
$$ \text{Speed of sound, } v = 339\,\mathrm{m/s} $$ $$ \text{Wavelength, } \lambda = 1.5\,\mathrm{cm} = 1.5 \times 10^{-2}\,\mathrm{m} $$To Find: Frequency of the sound (\( \nu \))
Using the wave equation:
$$ v = \lambda \nu $$ \[ \implies \nu = \frac{v}{\lambda} \]Substituting the given values:
$$ \begin{aligned} \nu &= \frac{339}{1.5 \times 10^{-2}} \\\\ \nu &= 22600\,\mathrm{Hz} \end{aligned} $$Conclusion:
The frequency of the sound wave is: $$ \boxed{\nu = 2.26 \times 10^{4}\,\mathrm{Hz}} $$Since the frequency is greater than \( 20{,}000\,\mathrm{Hz} \), it lies beyond the audible range of the human ear and is therefore classified as an ultrasound.
Q14. What is reverberation? How can it be reduced?
Solution:
Reverberation:
is the persistence of sound in a room or hall due to repeated
reflections
from walls, ceilings,
and other surfaces, even after the original sound has stopped.
When sound waves reflect many times before dying out, they overlap and make the sound seem prolonged or
blurred.
This effect is called reverberation. For example, in an empty hall or auditorium, your voice may sound
prolonged
because of multiple reflections.
Reduction of Reverberation:
Reverberation can be reduced by absorbing the reflected sound using soft and porous materials. Some
effective
ways include:
- Using sound-absorbing materials such as carpets, curtains, and thick cloth covers.
- Placing soft seats and wall panels made of cork, foam, or acoustic tiles.
- Installing false ceilings or wooden panels to absorb and scatter sound reflections.
- Adding wall hangings or soft boards in halls and studios.
Q15. What is the loudness of sound? What factors does it depend on?
Solution:
Loudness is the sensation of sound intensity as experienced by our ears. It
tells
us how strong or weak a sound
appears to us.
Two sounds may carry the same amount of energy, but one can seem louder than the other, depending on how our
ear
perceives it. For example, a drum sounds louder than a flute, even if both are played at the same
distance.
Factors on which Loudness depends:
- Amplitude of vibration:
Loudness mainly depends on the amplitude of the vibrating source. A sound wave with a larger amplitude carries more energy and is heard as louder, while a smaller amplitude produces a softer sound. - Sensitivity of the ear:
Different people may hear the same sound differently depending on the sensitivity of their ears. - Distance from the source:
Loudness decreases as we move farther away from the sound source because the energy spreads over a larger area. - Medium of propagation:
The medium through which sound travels (air, water, solid) also affects how loud a sound appears.
Q16. How is ultrasound used for cleaning?
Solution:
Ultrasound cleans objects by creating microscopic bubbles that dislodge dirt through high-frequency
vibrations —
a process called cavitation.
Q17. Explain how defects in a metal block can be detected using ultrasound.
Solution:
Ultrasound can detect defects inside metal blocks by using the principle of reflection of sound
waves.
Ultrasonic waves are sent into the metal block from one side using an ultrasonic detector. Normally, these
waves
travel through the metal and are reflected back from the opposite surface.
If there is a crack or air gap inside the metal, the waves get reflected earlier from that defect. By
comparing
the time or pattern of the reflected waves, the position and size of the defect can be identified.
