CBSE Class 7 Science Advanced Physics Sound Long Answer Questions Set 01

Long Answer Questions

Question 1. Explain how the sound produced by a source reaches the ear with a neat diagram.
Answer:
1. A vibrating source (for example a tuning fork or a vocal cord) pushes and pulls on the air next to it, forming alternate compressions (high pressure) and rarefactions (low pressure).
2. These pressure variations move through the air as a longitudinal wave; the air particles only vibrate to and fro and pass the disturbance on.
3. The wave enters the ear canal and makes the ear drum vibrate.
4. The vibrations pass through the small hinged bones of the middle ear to the inner ear.
5. The auditory nerve carries the signals to the hearing centre of the brain, and we hear the sound.

In simple words: Source, compressions and rarefactions in air, ear drum, bones, nerve, brain.

Teacher's Note:
1. Draw the ear diagram and label the sound waves, ear drum and the bones.
2. Write the path in order, like a flow of steps.

Question 2. Explain how sound is produced in humans.
Answer:
1. Sound in humans is produced by the vibration of the vocal cords, which are stretched across the larynx with a narrow slit (gap) between them.
2. Air from the lungs is forced through the slit and makes the cords vibrate.
3. The frequency depends on the thickness and tension of the cords: tight, thin cords give a high frequency (shrill sound), loose, thick cords give a low frequency sound.
4. Muscles attached to the cords control the tension and hence the pitch.
Activity: stretch two strips of rubber band one above the other and blow between them; stretching them more raises the pitch, and thinner strips give a still higher pitch.
In simple words: Air from the lungs vibrates the vocal cords and their tightness fixes the pitch.

Teacher's Note:
1. Link thickness and tension to the frequency.
2. The rubber band model is easy to try at home.

Question 3. What are the different measures to be taken to reduce noise pollution?
Answer:
1. Fit silencers in automobiles.
2. Reduce the use of loud speakers.
3. Locate factories away from residential areas.
4. Play the television, radio and music player at a low volume.
5. (Additional) Plant trees and use sound absorbing material in noisy places.
In simple words: Make less noise, place noisy things far from homes and absorb sound.

Teacher's Note:
1. The first four points are the standard ones from the chapter.
2. Point 5 is an extra practical measure.

Question 4. Describe an activity to show that pitch depends on the thickness of the vibrating body.
Answer:
1. Cut a rubber band into two pieces of equal length, lay one over the other, stretch them and blow air between them. Note the pitch of the sound.
2. Stretch the same pieces more and blow again; the pitch rises.
3. Now take a thin rubber band, cut it into two equal pieces, stretch and blow air between them as before.
4. The pitch of the sound is much higher with the thin rubber band.
Conclusion: the thinner the vibrating body, the higher the frequency and the higher the pitch.
In simple words: A thin rubber band sounds shriller than a thick one.

Teacher's Note:
1. Keep the length and the stretch the same when you compare.
2. The same reasoning explains why thin guitar strings give higher notes.

Question 5. Explain the propagation of sound in air.
Answer:
1. A vibrating object moves forward and compresses the air in front of it, creating a region of high pressure called a compression (C).
2. When the object moves back it leaves a region of low pressure called a rarefaction (R).
3. As the object vibrates rapidly, a series of compressions and rarefactions is created and moves through the air away from the source.
4. The air particles vibrate to and fro parallel to the direction in which the wave travels, so a sound wave is longitudinal.
5. Pressure is related to density, so sound propagation is a propagation of pressure (density) variations.

In simple words: A vibrating source makes alternate squeezed and stretched layers of air that travel outward.

Teacher's Note:
1. Define compression and rarefaction clearly.
2. Say that the particles do not travel with the wave; they only vibrate.

Question 6. What are wind and percussion instruments? Give examples.
Answer:
Wind (reed) instruments: long hollow pipes with a number of holes on the sides. Blowing air into the pipe makes the air column inside vibrate and produce musical sound. Examples: flute, bassoon, oboe.
Percussion (drum) instruments: they have stretched membranes of leather. When the membrane is struck it vibrates and gives musical sound; different notes are produced by changing the tension. Examples: drum, mridangam, tabla.
In simple words: Wind instruments have a vibrating air column, percussion instruments have a vibrating stretched skin.

Teacher's Note:
1. Say what vibrates in each type.
2. Give at least two examples each.

Question 7. (a) What is sonography?
(b) How do bats and whales make use of ultrasonic sounds?
(c) What are the harmful effects of noise pollution?
Answer:
(a) Sonography is the technique of using ultrasound waves to get images of the internal organs of the body, for example to see the condition of a baby inside the womb.
(b) Bats send out ultrasonic sound; it is reflected from insects and the bat finds its prey from the returning echo. Whales and dolphins use ultrasound in the same way to locate prey.
(c) Noise pollution causes loss of hearing, loss of work efficiency and mental tension (stress).
In simple words: Scanning with ultrasound, finding prey from echoes, and the damage that noise does.

Teacher's Note:
1. Answer each part under its own letter.
2. Bats and whales use echo location.

Question 8. (a) What are the higher units of frequency?
(b) How is noise pollution determined at a given place?
(c) Name some sound absorbing materials.
Answer:
(a) Kilohertz (kHz) and megahertz (MHz): 1 kHz = 1000 Hz and 1 MHz = 10⁶ Hz.
(b) It is determined by the loudness of the sound and the duration for which it lasts at that place.
(c) Cloth, curtains, thermocol, gunny bags, carpets, cork and other soft porous materials.
In simple words: Units kHz and MHz; loudness and duration judge noise; soft materials absorb sound.

Teacher's Note:
1. The book says noise pollution depends on loudness and duration.
2. Absorbers are soft and porous.

Question 9. Describe an experiment to show that sound cannot propagate in vacuum.
Answer:
1. Hang an electric bell in a glass jar closed with a cork lid; the jar has an outlet near the bottom connected to a vacuum pump. Switch on the bell; its sound is heard.
2. Start the vacuum pump and slowly remove the air from the jar.
3. As the air is removed the sound grows fainter and finally cannot be heard, though the bell can still be seen working.
Conclusion: sound needs a material medium; it cannot travel through a vacuum.

In simple words: The bell becomes silent when the air is pumped out, so sound needs a medium.

Teacher's Note:
1. Mention that the clapper is still seen moving.
2. Draw and label the bell jar apparatus.

Question 10. Sandeep constructed a simple pendulum to find the value of ‘g’ at that place. He observed that the bob of a simple pendulum moves from the mean position to one of the extreme positions in 0.5 s. Using this he determined the (i) time period of the simple pendulum, (ii) number of oscillations the pendulum has undergone in four seconds and (iii) the frequency of oscillation of the pendulum. What are his answers?
Answer:
(i) Mean to one extreme is one quarter of an oscillation, so T/4 = 0.5 s and T = 2 s.
(ii) Number of oscillations in 4 s = 4 ÷ 2 = 2.
(iii) f = 1/T = 1/2 = 0.5 Hz.
In simple words: Mean to extreme is T/4, so T = 2 s, two oscillations in 4 s, f = 0.5 Hz.

Teacher's Note:
1. Remember that mean to extreme is only a quarter of an oscillation.
2. T = 2 s makes it a seconds pendulum.

Question 11. Chanakya was eager to know the time taken by sound to travel the same distance in water and air. So, he set a body on vibration with a frequency of 250 Hz in air and the sound produced by this is heard at a distance of 175 m. Now if the body is vibrated in water, what is the difference in the time taken by the sound to travel the same distance in water and air? (The velocity of sound in air and water is 350 m s⁻¹ and 1400 m s⁻¹, respectively.)
Answer:
Time in air: t₁ = 175 ÷ 350 = 1/2 s = 0.5 s.
The frequency of the body (250 Hz) remains the same in water.
Time in water: t₂ = 175 ÷ 1400 = 1/8 s = 0.125 s.
Difference = 1/2 − 1/8 = 4/8 − 1/8 = 3/8 s = 0.375 s.
In simple words: Sound takes 0.5 s in air and 0.125 s in water, a difference of 0.375 s.

Teacher's Note:
1. The frequency does not affect the travel time.
2. Find each time separately and subtract.

Question 12. The velocity of sound in air, in water and in a given solid is, 330 m s⁻¹, 1350 m s⁻¹ and 1500 m s⁻¹, respectively. If a vibrating tuning fork of frequency 150 Hz is placed in each of these media, then find the distance covered by the sound in air, water and solid, respectively, at the end of 50 vibrations.
Answer:
Time for 50 vibrations = 50 ÷ 150 = 1/3 s.
Distance in air = 330 × 1/3 = 110 m.
Distance in water = 1350 × 1/3 = 450 m.
Distance in solid = 1500 × 1/3 = 500 m.
In simple words: Find the time of 50 vibrations, then multiply by each speed.

Teacher's Note:
1. The frequency stays 150 Hz in all three media.
2. Distance = speed × time.

Question 13. The audible range of frequency is 20 Hz to 20,000 Hz. Determine the corresponding range of time periods of vibrations.
Answer:
T = 1/f.
For 20 Hz: T₁ = 1/20 = 0.05 s = 5 × 10⁻² s.
For 20 000 Hz: T₂ = 1/20 000 = 0.00005 s = 5 × 10⁻⁵ s.
The range of time periods is 5 × 10⁻² s to 5 × 10⁻⁵ s.
In simple words: Take the reciprocal of both end frequencies.

Teacher's Note:
1. The low frequency has the long time period.
2. Check that f × T = 1.

Question 14. A bat entering into a well finds it difficult to come out of it. Explain the reason.
Answer:
A bat finds its way by sending out ultrasonic sound and listening to the echo.
Inside a well the ultrasound is reflected from all sides, so the bat receives echoes from every direction and cannot find the path out.
In simple words: Echoes from all walls confuse the bat.

Teacher's Note:
1. Bats use echo location instead of sight.
2. Many echoes at once give no clear direction.

Question 15. A student breaks a fused bulb, and then hears a sound of explosion. What could be the reason for this? Explain.
Answer:
Inside a fused bulb there is only a little low-pressure inert gas.
When the bulb is broken, air from outside rushes in suddenly to fill the space and strikes the gas inside.
This rapid movement of air creates a pressure wave that is heard as a sound like an explosion.
In simple words: Air rushes into the low-pressure bulb and makes a bang.

Teacher's Note:
1. Outside pressure is higher than the pressure inside.
2. Sudden compression of air gives a loud sound.

Question 16. How are dogs made alert by the crime department? Explain.
Answer:
Dogs can hear sounds of frequency greater than 20 kHz, which humans cannot hear.
The crime department uses a Galton whistle, which produces ultrasound. The dog hears it and is alerted, while people around do not hear anything.
In simple words: A Galton whistle makes ultrasound that only dogs can hear.

Teacher's Note:
1. The whistle works above the human range of 20 kHz.
2. Dogs respond to about 40 000 Hz.

Question 17. Three boys Ravi, John and Asif stand in a straight line as shown in the figure. Ravi and Asif fire crackers at the same instant of time. John hears the first sound after 2 s and the second sound after 2.5 s. Determine the distance between Ravi and Asif. The velocity of sound in air is 330 m s⁻¹.

Answer:
Distance from Ravi to John: d₁ = 330 × 2 = 660 m.
Distance from Asif to John: d₂ = 330 × 2.5 = 825 m.
John is between the two, so the distance between Ravi and Asif = 660 + 825 = 1485 m.
In simple words: Each distance is speed × time, and the two add up.

Teacher's Note:
1. Draw the three boys on a line first.
2. The nearer cracker is heard first.