Sound
NCERT Solutions • Class 9 Science • Chapter 11Chapter Exercises
1. What is sound and how is it produced?
Sound is a form of energy which produces a sensation of hearing in our ears.
Production: Sound is produced by vibrating objects. When an object vibrates (moves back and forth rapidly), it sets the particles of the medium around it in vibration, creating sound waves.
Production: Sound is produced by vibrating objects. When an object vibrates (moves back and forth rapidly), it sets the particles of the medium around it in vibration, creating sound waves.
2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
When a vibrating object (like a tuning fork) moves forward, it pushes and compresses the air in front of it, creating a region of high pressure called Compression (C).
When the vibrating object moves backward, it creates a region of low pressure called Rarefaction (R).
As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air, forming a sound wave.
When the vibrating object moves backward, it creates a region of low pressure called Rarefaction (R).
As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air, forming a sound wave.
3. Why is sound wave called a longitudinal wave?
Sound waves are called longitudinal waves because the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles vibrate back and forth about their position of rest.
4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
The characteristic is Timbre (or Quality). It allows us to distinguish between two sounds having the same pitch and loudness.
5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Speed of Light vs Sound
The speed of light ($3 \times 10^8 \text{ m/s}$) is exceedingly high compared to the speed of sound ($344 \text{ m/s}$ in air). Therefore, the flash of lightning reaches our eyes almost instantly, whereas the sound of thunder takes a few seconds to travel the same distance.
The speed of light ($3 \times 10^8 \text{ m/s}$) is exceedingly high compared to the speed of sound ($344 \text{ m/s}$ in air). Therefore, the flash of lightning reaches our eyes almost instantly, whereas the sound of thunder takes a few seconds to travel the same distance.
6. 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 $344 \text{ m s}^{-1}$.
Formula: $\lambda = v / f$
For $f = 20 \text{ Hz}$:
$\lambda_1 = \frac{344}{20} =$ $17.2 \text{ m}$.
For $f = 20 \text{ kHz} = 20000 \text{ Hz}$:
$\lambda_2 = \frac{344}{20000} =$ $0.0172 \text{ m}$.
For $f = 20 \text{ Hz}$:
$\lambda_1 = \frac{344}{20} =$ $17.2 \text{ m}$.
For $f = 20 \text{ kHz} = 20000 \text{ Hz}$:
$\lambda_2 = \frac{344}{20000} =$ $0.0172 \text{ m}$.
7. 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.
Let length of rod be $d$.
Speed in Air ($v_{air}$) $\approx 344 \text{ m/s}$.
Speed in Aluminium ($v_{Al}$) $\approx 6420 \text{ m/s}$.
Time taken in Air, $t_{air} = \frac{d}{v_{air}}$.
Time taken in Al, $t_{Al} = \frac{d}{v_{Al}}$.
Ratio: $\frac{t_{air}}{t_{Al}} = \frac{d/v_{air}}{d/v_{Al}} = \frac{v_{Al}}{v_{air}}$
$= \frac{6420}{344} \approx$ $18.66$.
Speed in Air ($v_{air}$) $\approx 344 \text{ m/s}$.
Speed in Aluminium ($v_{Al}$) $\approx 6420 \text{ m/s}$.
Time taken in Air, $t_{air} = \frac{d}{v_{air}}$.
Time taken in Al, $t_{Al} = \frac{d}{v_{Al}}$.
Ratio: $\frac{t_{air}}{t_{Al}} = \frac{d/v_{air}}{d/v_{Al}} = \frac{v_{Al}}{v_{air}}$
$= \frac{6420}{344} \approx$ $18.66$.
8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Frequency = $100 \text{ Hz}$ (100 vibrations per second).
Time = $1 \text{ minute} = 60 \text{ seconds}$.
Total Vibrations = $100 \times 60 =$ $6000 \text{ times}$.
Time = $1 \text{ minute} = 60 \text{ seconds}$.
Total Vibrations = $100 \times 60 =$ $6000 \text{ times}$.
9. Does sound follow the same laws of reflection as light does? Explain.
Yes. Sound follows the same laws of reflection as light:
1. The angle of incidence is equal to the angle of reflection.
2. The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.
1. The angle of incidence is equal to the angle of reflection.
2. The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.
10. 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?
Explanation:
To hear an echo, the time interval between the original sound and reflected sound must be at least 0.1 s.
On a hotter day, the temperature increases. As temperature increases, the speed of sound increases.
Since $Time = \frac{Distance}{Speed}$, if speed increases, time decreases.
If the time interval becomes less than 0.1 s due to higher speed, you may NOT hear the echo on a hotter day, even if you heard it on a cooler day.
To hear an echo, the time interval between the original sound and reflected sound must be at least 0.1 s.
On a hotter day, the temperature increases. As temperature increases, the speed of sound increases.
Since $Time = \frac{Distance}{Speed}$, if speed increases, time decreases.
If the time interval becomes less than 0.1 s due to higher speed, you may NOT hear the echo on a hotter day, even if you heard it on a cooler day.
11. Give two practical applications of reflection of sound waves.
1. Megaphones/Horns: Designed to send sound in a particular direction by multiple reflections.
2. Stethoscope: Used by doctors to listen to sounds within the body (heartbeat) via multiple reflections of sound in the tube.
2. Stethoscope: Used by doctors to listen to sounds within the body (heartbeat) via multiple reflections of sound in the tube.
12. A stone is dropped from the top of a tower 500 m high… When is the splash heard at the top? ($g = 10 \text{ m s}^{-2}, v_{sound} = 340 \text{ m s}^{-1}$)
Step 1 Time for stone to fall ($t_1$):
$s = ut + \frac{1}{2}gt^2$
$500 = 0 + \frac{1}{2}(10)t_1^2 \Rightarrow 500 = 5t_1^2 \Rightarrow t_1^2 = 100$
$t_1 = 10 \text{ s}$.
Step 2 Time for sound to return ($t_2$):
$t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{500}{340} \approx 1.47 \text{ s}$.
Total Time
$t = t_1 + t_2 = 10 + 1.47 =$ $11.47 \text{ s}$.
$s = ut + \frac{1}{2}gt^2$
$500 = 0 + \frac{1}{2}(10)t_1^2 \Rightarrow 500 = 5t_1^2 \Rightarrow t_1^2 = 100$
$t_1 = 10 \text{ s}$.
Step 2 Time for sound to return ($t_2$):
$t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{500}{340} \approx 1.47 \text{ s}$.
Total Time
$t = t_1 + t_2 = 10 + 1.47 =$ $11.47 \text{ s}$.
13. A sound wave travels at a speed of $339 \text{ m s}^{-1}$. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Given: $v = 339 \text{ m/s}$, $\lambda = 1.5 \text{ cm} = 0.015 \text{ m}$.
Frequency ($f$): $v / \lambda = 339 / 0.015$
$f =$ $22600 \text{ Hz}$.
Audibility: No, it will not be audible because it is ultrasonic (Frequency > 20,000 Hz). The audible range is 20 Hz – 20,000 Hz.
Frequency ($f$): $v / \lambda = 339 / 0.015$
$f =$ $22600 \text{ Hz}$.
Audibility: No, it will not be audible because it is ultrasonic (Frequency > 20,000 Hz). The audible range is 20 Hz – 20,000 Hz.
14. What is reverberation? How can it be reduced?
Reverberation is the persistence of sound in a big hall due to repeated reflections from the walls, ceiling, and floor.
Reduction Methods:
1. Covering the roof and walls with sound-absorbent materials like compressed fibreboard.
2. Using rough plaster or heavy curtains.
3. Upholstering seats with sound-absorbing materials.
Reduction Methods:
1. Covering the roof and walls with sound-absorbent materials like compressed fibreboard.
2. Using rough plaster or heavy curtains.
3. Upholstering seats with sound-absorbing materials.
15. What is loudness of sound? What factors does it depend on?
Loudness is the physiological response of the ear to the intensity of sound. It distinguishes a loud sound from a faint one.
Factors: It depends primarily on the Amplitude of the vibration. Greater the amplitude, louder the sound.
Factors: It depends primarily on the Amplitude of the vibration. Greater the amplitude, louder the sound.
16. How is ultrasound used for cleaning?
Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, particles of dust, grease, and dirt detach and drop out. This method is used for spiral tubes, electronic components, etc.
17. Explain how defects in a metal block can be detected using ultrasound.
Ultrasounds can be used to detect cracks and flaws in metal blocks.
1. Ultrasonic waves are passed through the metal block.
2. Detectors are placed on the other side.
3. If there is even a small defect or crack, the ultrasound gets reflected back and does not reach the detector.
4. The absence of the signal at the detector indicates the presence of a flaw.
1. Ultrasonic waves are passed through the metal block.
2. Detectors are placed on the other side.
3. If there is even a small defect or crack, the ultrasound gets reflected back and does not reach the detector.
4. The absence of the signal at the detector indicates the presence of a flaw.