Chapter 11 -Sound
Best NCERT Solutions for Class 9 Science Chapter 11 Sound .A vital component of our daily lives, sound shapes our interactions with the outside world and impacts both our mental and physical health. This chapter will examine the physics and importance of sound, covering everything from its fundamental physical characteristics to its intricate function in human perception and communication.
Exercise 1
1. How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
The adjacent particles in the medium vibrate in response to an object’s vibrations. The nearby particles receive additional vibrations from these. same ways that these vibrations travel between particles before arriving at our ears.
Exercise 2
1. Explain how sound is produced by your school bell.
Answer:
The nearby airborne particles are compelled to vibrate simultaneously when the school bell rings. A sound wave is created as a result of this disruption, and as the bell advances, the air in front of it is forced forward. As a result, compression—a zone of high pressure—is created. Rarefaction is the term for the area of low pressure that results from the bell moving rearward.
The bell keeps moving forward and backward in this manner, causing a number of compressions and rarefactions. When it travels through the air, this produces the sound of a bell.
2. Why are sound waves called mechanical waves?
Answer:
Particles in close proximity vibrate as sound waves travel through a material. Through a series of compressions and rarefactions, particles in the medium interact to produce sound waves. These waves are hence referred to as mechanical waves.
3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: Since sound waves are mechanical in nature, they require a medium to spread. Since the moon lacks an atmosphere, we are unable to hear sound there.
Exercise 3
1. Which wave property determines (a) loudness, (b) pitch?
Answer:
(A) A sound’s amplitude determines how loud it is. The sound is louder when the amplitude is higher.
(b) A sound’s frequency determines its pitch. The pitch of sound increases with frequency.
2. Guess which sound has a higher pitch: guitar or car horn?
Answer: Compared to an automobile horn, the vibration frequency produced by a guitar is higher. The sound’s pitch is determined by its frequency. Pitch rises with increasing frequency. In other words, the guitar has a higher pitch than an automobile horn.
Exercise 4
1. What are the wavelength, frequency, time period and amplitude of a sound wave?
Answer:
The definitions that follow are provided below:
-Wavelength: The separation between two successive rarefactions or compressions is known as the wavelength. The metre (m) is its SI unit.
-The number of full oscillations per second is the definition of frequency for a sound wave. It is expressed in Hz, or hertz.
-Duration: The duration of a sound wave is the amount of time it takes for one cycle to finish. Seconds are its SI unit (s).
-The maximum extent of a vibration, measured from the equilibrium position, is the amplitude of a sound wave.
2. How are the wavelength and frequency of a sound wave related to its speed?
Answer: Speed, wavelength, and frequency of a sound wave are related by the equation given below:
∴v=λ×υ∴v=λ×υ
3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
It is stated in the question above that:
The sound wave’s frequency is υ=220 Hz.
The sound wave travels at v=440 m/s.
Regarding a wave of sound,
Wavelength (λ) x Frequency (υ) equals speed (v).
∴λ=440220=2 meters
Consequently, the sound wave’s wavelength is 2 m.
4. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer:
It is stated in the question above that:
500 Hz is the frequency.
The wave’s period is equal to the time interval between two subsequent compressions. As we are aware,
Duration = 1 frequency = 1500 = 0.002 s
Exercise 5
1. Distinguish between loudness and intensity of sound.
Answer: The quantity of sound energy that travels through a unit area in a second is known as the intensity of a sound wave. A sound’s amplitude mostly determines how loud or soft it is. The force that an object vibrates determines the amplitude of the sound wave. The amplitude of a sound wave, which the ear interprets as loudness, is determined in part by intensity.
Exercise 6
1. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
When it comes to solids, like iron, water, and air at a specific temperature, sound travels the fastest.
Exercise 7
1. An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer:
It is stated in the question above that:
The sound speed is v = 342 m/s.
Time-wise, echo returns in t = 3s.
The formula to find the distance travelled by sound is v × t = 342 × 3 = 1026 m.
The reflecting surface’s distance from the source, as the sound travels and is reflected back, is 10262 = 513 meters.
Exercise 9
1. What is the audible range of the average human ear?
Answer:
An average human ear can hear sounds in the 20–20,000 Hz range. Sounds having a frequency of less than 20 Hz are inaudible to humans.
and above 20,000 hertz
2. What is the range of frequencies associated with
(a) Infrasound?
Answer: The frequencies of infrasound are lower than 20 Hz.
(b) Ultrasound?
Answer: There are frequencies in ultrasound that exceed 20,000 Hz.
NCERT Exercise
1. What is sound and how is it produced?
Answer: Vibrations in the form of energy give rise to the sense of hearing, or sound. An object’s vibrations propagate to the nearby particles in the medium, causing those particles to vibrate as well. Waves are produced in the medium as a result of this disruption. Thus, sound is produced when this disruption gets to the ear.
2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
Air is the most frequent medium for sound to move through. A zone of high pressure is created when an object that is vibrating advances ahead and pushes and compresses the surrounding air. We refer to this area as a compression (C). The compression begins to recede from the thing that is vibrating.
The backward motion of the vibrating object generates rarefaction (R), a low-pressure area. A sequence of compressions and rarefactions are produced in the air as the object rapidly oscillates back and forth. The sound wave that travels through the medium is created by them.
Increased particle density in the medium results in increased pressure, and vice versa. As a result, one way to depict the transmission of sound is as fluctuations in pressure or density inside the medium.
3. Why is a sound wave called a longitudinal wave?
Answer:
When it comes to sound waves, the particles only oscillate back and forth around their rest state rather than moving. The individual medium particles in sound waves travel in a direction parallel to the disturbance’s direction of propagation. Thus, a sound wave is a longitudinal wave.
4. Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality or timber of sound that allows us to distinguish one sound from another having the same pitch and loudness is the characteristic of the sound that helps you identify your friend by his voice when you’re seated among others in a dark room.
5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Answer:
Sound travels at 344 m/s, but light travels at 3 × 108 m/s. Thunder requires more time than light to travel to Earth because light travels faster than sound. Thus, we see a flash before we hear thunder.
6. A person has a hearing range from
to
. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as
.
Answer:
It is stated in the question above that:
An individual’s hearing range starts at 20 Hz.
up to 20 kHz
In air, the speed of sound is 344 m/s.
We are aware of:
In the case of υ1=20 Hz, λ1=vυ1=34420=17.2m; in the case of υ2=20 kHz, λ2=vυ2=34420000=0.0172m. Speed is equal to Wavelength × Frequency.
Therefore, the wavelength range in which humans can hear is 0.0172 m to 17.2 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 time taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Think about the aluminium rod’s required length, d.
Sound wave speed in metal at 25°C
vAl equals 6420 m/s.
Consequently, the duration required for the sound wave to travel to the opposite end is:
tAl=dvAl=d6420
Sound wave speed in air at 25°C
vAir equals 346 meters per second.
As a result, the duration of a sound wave to travel to the opposite end is:
tAir=dvAir=d346
As a result, the ratio of the time the sound wave takes in metal to that in air is:
tAirtAl=6420346=d346d6420=18.55
You guys can also follow us on the social media
8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
It is stated in the question above that:
The sound frequency is 100 Hz.
Total time: 60 seconds/1 minute
We are aware that the number of oscillations per second is the definition of frequency. It’s provided by the relationship:
Duration × Frequency = Number of oscillations
One oscillation is equal to 100 × 60 = 6000.
The source vibrates 6000 as a result.
resulting in a frequency of 100 Hz each minute.
9. Does sound follow the same laws of reflection as light does? Explain.
Answer: At the point of incidence, the incident and reflected sound waves form the same angle with respect to the surface normal. Furthermore, the plane in which the incident, reflected, and normal sound waves are all located is the same as the point of incidence. Thus, sound and light are subject to the same rules of reflection.
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 remain the same. Do you hear the echo sound on a hotter day?
Answer:
When there is a minimum of 0.1 seconds between the source sound and the reflected sound, an echo is audible. The temperature of a medium has a direct relationship with sound speed.
As a result, the temperature and time interval will be directly inversely proportional. As a result, on a hotter day, the time delay between the original sound and the sound that is reflected will be shorter.
11. Give two practical applications of reflection of sound waves.
Answer: Following are the two practical applications of reflection of sound waves:
(a) SONAR:
SONAR is a technology that uses sound reflection to gauge an underwater object’s speed and distance.
(b) Stethoscope:
With the use of a stethoscope, a doctor may hear a patient’s heartbeat through various reflections of sound.
12. 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,
and
.
Answer:
It is stated in the question above that:
The tower’s height is s = 500 meters.
The sound’s velocity is v = 340 m/s.
Gravity-related acceleration is g = 10 m/s2.
The stone’s initial velocity will be u = 0 m/s since it is initially at rest.
Let t1 be the amount of time it took for the stone to fall to the tower’s base.
The second equation of motion states that:
s = ut1 + 12gt12 ⇒500= 12(10)(t1)2 ⇒ (t1)2 = 100 ⇒ t1 = 10s
From the base of the tower, the sound will now take the following amount of time to reach the top:
⇒t2=500340=1.47 seconds
Therefore, at time t=t1+t2=10+1.47=11.47s, the splash is audible at the top.
13. A sound wave travels at a speed of
. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
It is stated in the question above that:
The sound travels at 339 m/s.
The sound wavelength is λ = 1.5 cm = 0.015 m.
We are aware of:
Wavelength × Frequency = Speed of Sound ⇒v=λ×υ
Thus, frequency υ
shall be:
⇒υ=vλ=22600Hz/3390.015
Given that humans can perceive sound in a frequency range of 20 Hz to
up to 20,000 Hz
Given that the sound’s frequency exceeds 20,000 Hz,
won’t be able to be heard.
14. What is reverberation? How can it be reduced?
Answer:
Reverberation is the continuation of sound after the source ceases to produce sound as a result of multiple reflections. In a large hall, sound waves bounce off the walls repeatedly, reducing the intensity of the sound until it is undetectable.
The auditorium’s walls and roof are often covered with rough plaster, compressed fibreboard, or drapes, or other sound-absorbing materials, to lessen reverberation. The materials used to make the seats are chosen for their ability to absorb sound.
15. What is the loudness of sound? What factors does it depend on?
Answer:
The loudness of the sound is the measurement of the ear’s response to it.
The force used to cause an object to vibrate determines the amplitude of sound, which in turn determines how loud or soft it is. A loud noise has a lot of energy.
The amplitude of vibrations determines loudness, and loudness is inversely correlated with the square of the vibration amplitude.
16. How is ultrasound used for cleaning?
Answer: When cleaning, ultrasonic waves are directed through the items submerged in the cleaning solution. The things are cleaned of grime by their high frequency.
17. Explain how defects in a metal block can be detected using ultrasound.
Answer:
Large-scale construction projects like buildings, bridges, machinery, and scientific apparatus typically require metallic components. Metal blocks can have imperfections and cracks found with ultrasounds.
The structure’s strength is decreased by internal cracks or holes in the metal blocks that are not visible from the outside. The metal block permits ultrasonic waves to flow through it, and the transmitted waves are detected by detectors.
The ultrasonic wave is reflected back if there is even a slight imperfection. This suggests that there is a fault or other imperfection present.
Best NCERT Solutions for Class 9 Science Chapter 11 Sound