Class 9 Science Chapter 11 Sound

Key Concept: Infrasonic Sound

a) 10 Hz

[Solution Description]

Infrasonic sound refers to sound waves with frequencies below 20 Hz.

Key Concept: Reflection of Sound – Megaphones and Loudspeakers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that megaphones use multiple reflections of sound to amplify it, which is true because megaphones are designed with a tube and a conical opening that reflect sound waves successively, guiding them in a forward direction. This process increases the intensity of the sound. The reason explains that multiple reflections guide sound waves in a specific direction, which is also true and directly explains why megaphones amplify sound. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Echolocation Principle

d) They reflect back from defective locations without bending

[Solution Description]

Ultrasonic waves have a higher frequency and shorter wavelength compared to ordinary sound waves. This allows them to travel along well-defined paths even in the presence of obstacles. They can detect cracks and flaws in metal blocks because they reflect back from defective locations without bending around corners, unlike ordinary sound waves which can bend around corners and interfere with detection.

Key Concept: Sound Reflection in Concert Halls

c) To reflect sound evenly across the entire hall

[Solution Description]

The curved ceilings in concert halls are designed to reflect sound waves evenly across the entire hall. This ensures that sound reaches all corners of the hall, providing a uniform listening experience for the audience. The curvature helps in distributing the sound more effectively compared to flat ceilings.

Key Concept: Application of Ultrasound

c) Vibration caused by high-frequency sound waves

[Solution Description]

Ultrasound breaks kidney stones into fine grains through a process called lithotripsy. High-frequency sound waves are focused on the stones, creating vibrations that cause the stones to fracture into smaller pieces. These smaller grains can then be naturally flushed out of the body with urine.

Key Concept: Medical Uses of Ultrasound

c) Examining the foetus

[Solution Description]

Ultrasonography is used during pregnancy to examine the foetus and detect any congenital defects or growth abnormalities.

Key Concept: Purpose of Ultrasonography

c) To detect congenital defects and growth abnormalities in the foetus

[Solution Description]

During pregnancy, ultrasonography is primarily used to examine the foetus to detect congenital defects and growth abnormalities.

Key Concept: Ultrasound Frequency Range

a) Above 20000 Hz

[Solution Description]

Ultrasound refers to sound waves with frequencies above 20000 Hz.

Key Concept: Medical Applications of Ultrasound

b) Echocardiography

[Solution Description]

Echocardiography is a technique where ultrasonic waves are reflected from various parts of the heart to form an image of the heart.

Key Concept: Reflection of Sound

c) The angle of incidence is equal to the angle of reflection.

[Solution Description]

The law of reflection of sound states that the directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface at the point of incidence, and the three lie in the same plane. This means the angle of incidence is equal to the angle of reflection, and both the incident and reflected rays lie in the same plane as the normal.

Key Concept: Reflection of Sound, Laws of Reflection

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that a polished surface is required for effective reflection of sound waves. However, according to the syllabus, an obstacle of large size, which may be polished or rough, is needed for the reflection of sound waves. This means that while a polished surface can enhance reflection by minimizing scattering, it is not strictly necessary. The reason claims that polished surfaces minimize the scattering of sound waves, which is true as they provide a smoother surface for reflection. However, this does not fully explain why a polished surface is required, as reflection can also occur on rough surfaces. Therefore, the assertion is false, but the reason is true.

Key Concept: Reverberation

b) Reverberation

[Solution Description]

The persistence of sound in an auditorium due to repeated reflections of sound is called reverberation.

Key Concept: Reflection of Sound

c) The sound wave reflects at an angle equal to the angle of incidence

[Solution Description]

Sound waves follow the laws of reflection, where the angle of incidence equals the angle of reflection, and both lie in the same plane as the normal to the reflecting surface. A polished wall reflects sound waves efficiently, causing them to bounce off at equal angles.

Key Concept: Echo formation, Temperature effect on speed of sound

b) Increases by 0.3 m

[Solution Description]

The minimum distance required to hear a distinct echo is directly proportional to the speed of sound. As the speed of sound increases, the minimum distance also increases. Given the initial speed $v_1 = 343 \, \text{m/s}$ and the new speed $v_2 = 349 \, \text{m/s}$, we calculate the minimum distance at each speed using the formula $d = (v \times 0.1)/2$. At $20^\circ \text{C}$, the minimum distance is $(343 \times 0.1)/2 = 17.15 \, \text{m}$. At $30^\circ \text{C}$, the minimum distance is $(349 \times 0.1)/2 = 17.45 \, \text{m}$. Therefore, the minimum distance increases by $17.45 – 17.15 = 0.3 \, \text{m}$.

Key Concept: Echo, Speed of Sound

a) 680 m

[Solution Description]

To find the distance between the person and the mountain, we use the formula for the distance traveled by sound: $d = v \times t$, where $v$ is the speed of sound and $t$ is the time taken for the echo to return. Here, the sound travels to the mountain and back, so the total distance is twice the distance between the person and the mountain. Given that the time taken $t$ is 4 seconds and the speed of sound $v$ is 340 m/s, the total distance traveled by sound is $340 \, \text{m/s} \times 4 \, \text{s} = 1360 \, \text{m}$. Therefore, the distance between the person and the mountain is $1360 \, \text{m} / 2 = 680 \, \text{m}$.

Key Concept: Speed of Sound

b) 340 m/s

[Solution Description]

The speed of a sound wave can be calculated using the formula $v = \lambda\nu$, where $v$ is the speed, $\lambda$ is the wavelength, and $\nu$ is the frequency. Given that the frequency $\nu$ is 500 Hz and the wavelength $\lambda$ is 0.68 m, we can substitute these values into the formula:

$v = 0.68 \, \text{m} \times 500 \, \text{Hz} = 340 \, \text{m/s}$

Therefore, the speed of the sound wave is 340 m/s.

Key Concept: Longitudinal Waves

c) Longitudinal wave

[Solution Description]

Sound waves are longitudinal waves because the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles oscillate back and forth about their position of rest.

Key Concept: Infrasonic Sounds

c) Below 20 Hz

[Solution Description]

Infrasonic sounds are defined as sounds with frequencies below 20 Hz. This is a basic concept from the syllabus.

Key Concept: Echocardiography

b) To create images of the heart

[Solution Description] Echocardiography is a technique that uses ultrasonic waves to create images of the heart. The primary purpose of this technique is to visualize the internal structure and function of the heart, allowing doctors to detect abnormalities such as structural defects or functional issues.

Key Concept: Reverberation and Sound Absorption

c) Draperies

[Solution Description] To reduce reverberation, sound-absorbent materials are used. Materials like compressed fibreboard, rough plaster, and draperies are effective in absorbing sound and reducing its reflection. Among the options, draperies are particularly effective as they absorb sound waves more efficiently compared to hard surfaces like glass or concrete.

Key Concept: Successive Reflections

c) 525 m

[Solution Description]

The time difference between the first and second echoes is due to the additional distance traveled by the sound between the two cliffs. Let the distance between the two cliffs be $D$. The sound travels this distance twice (to the second cliff and back), so the total distance is $2D$. Using the formula $d = v \times t$, we have $2D = 350 \times 3 = 1050 \, \text{m}$. Therefore, $D = \frac{1050}{2} = 525 \, \text{m}$.

Key Concept: Ultrasonic Applications

c) Echocardiography

[Solution Description]

Ultrasound has several medical applications, including imaging internal organs and detecting abnormalities like stones or tumors. One such application is echocardiography, which uses ultrasonic waves to form images of the heart.

Key Concept: Pitch, Frequency, Speed of Sound

b) 0.68 m

[Solution Description]

To find the wavelength, we use the formula $v = \lambda \nu$, where $v$ is the speed of sound, $\lambda$ is the wavelength, and $\nu$ is the frequency. Given $v = 340$ m/s and $\nu = 500$ Hz, we can solve for $\lambda$:

$\lambda = \frac{v}{\nu} = \frac{340}{500} = 0.68 \text{ m}$

Therefore, the wavelength is 0.68 meters.

Key Concept: Process of Ultrasound Cleaning

a) Particles of dust, grease, and dirt get detached

[Solution Description]

When ultrasonic waves are sent into a cleaning solution, the high-frequency vibrations cause particles of dust, grease, and dirt to detach from the object being cleaned, resulting in thorough cleaning.

Key Concept: Need for a Medium

b) To create a disturbance that travels through particles

[Solution Description]

Sound requires a medium to travel because it involves the vibration of particles within that medium. These vibrations create compressions and rarefactions, which propagate as sound waves. Without a medium, there are no particles to vibrate, and thus sound cannot travel. This principle applies to solids, liquids, and gases, but not in a vacuum where there are no particles.

Key Concept: Law of Reflection of Sound

c) 55$^\circ$

[Solution Description] The angle of reflection is always equal to the angle of incidence. Rotating the surface changes the normal line by 10$^\circ$. Therefore, the new angle of incidence becomes $45^\circ + 10^\circ = 55^\circ$, and hence, the angle of reflection will also be 55$^\circ$.

Key Concept: Ultrasound Reflection

c) The ultrasound wave gets reflected back from the crack.

[Solution Description]

When an ultrasound wave encounters a crack or defect inside a metal block, it gets reflected back. This reflection occurs because the crack disrupts the continuous medium of the metal, causing the wave to bounce back to the detector. This behavior is due to the change in acoustic impedance at the crack interface.

Key Concept: Application of ultrasound

b) They can travel along well-defined paths and clean hard-to-reach areas.

[Solution Description] Ultrasonic waves are preferred for cleaning spiral tubes and odd-shaped machine parts because they can travel along well-defined paths and penetrate intricate geometries. Their high frequency allows them to reach areas that are difficult to clean using traditional methods. The cavitation process ensures that dirt and grease are removed uniformly from all surfaces, regardless of the shape or complexity of the object.

Key Concept: Law of Reflection of Sound

d) $40^\circ$

[Solution Description]

According to the problem, the angle of incidence is twice the angle of reflection. Given that the angle of reflection is $20^\circ$, the angle of incidence can be calculated as $2 \times 20^\circ = 40^\circ$. This satisfies the condition given in the problem, even though it contradicts the standard law of reflection where the angle of incidence equals the angle of reflection.

Key Concept: Sound Production by Vibrating Objects

b) Ripples are formed on the water surface

[Solution Description]

When a vibrating tuning fork touches the surface of water, it causes ripples to form on the water surface. This happens because the vibrating prong transfers its energy to the water molecules, setting them into motion. The rapid back-and-forth movement of the tuning fork creates disturbances in the water, which propagate as ripples. These ripples are a visual representation of the sound waves produced by the tuning fork.

Key Concept: Reflection of Sound

b) To amplify the sound by guiding it through reflections

[Solution Description]

The conical opening in a stethoscope helps in directing the sound waves more effectively towards the doctor’s ears. It aids in amplifying the sound by guiding the sound waves through multiple reflections, ensuring that most of the sound reaches the listener.

Key Concept: Ultrasound Frequency

c) They cause vibrations at high frequencies

[Solution Description] Ultrasonic waves are effective for cleaning because their high frequency causes vibrations that dislodge dirt, grease, and dust particles from the surfaces of objects placed in a cleaning solution.

Key Concept: Speed of sound in different states

c) Aluminium

[Solution Description]

From Table 11.1, the speed of sound in aluminium is $6420 \, \text{m/s}$, which is higher than the speed of sound in nickel ($6040 \, \text{m/s}$), steel ($5960 \, \text{m/s}$), and iron ($5950 \, \text{m/s}$). Therefore, aluminium has the highest speed of sound at $25 \, ^\circ\text{C}$.

Key Concept: Speed of Sound

c) Amplitude of the sound wave

[Solution Description]

The speed of sound primarily depends on the nature and temperature of the medium. It does not depend on the amplitude of the sound wave.

Key Concept: Sound Propagation

b) Particles vibrate but do not move forward

[Solution Description]

When sound propagates through a medium, the particles of the medium do not move forward with the sound. Instead, they vibrate about their equilibrium positions. The disturbance created by the sound wave causes each particle to exert a force on its neighboring particles, displacing them temporarily. This process continues, allowing the sound wave to travel through the medium without the particles themselves moving long distances.

Key Concept: Vibration and Sound Production

c) The ball vibrates along with the prong

[Solution Description]

When the vibrating tuning fork is brought near a suspended table tennis ball, the vibration of the tuning fork causes the ball to move. This happens because the vibrating prong transfers its energy to the ball, causing it to oscillate. The movement of the ball confirms that sound is produced due to vibrations.

Key Concept: Speed of Sound

b) $0.2 \, \text{m}$

[Solution Description]

The relationship between the speed of sound ($v$), frequency ($\nu$), and wavelength ($\lambda$) is given by the equation $v = \lambda \nu$. Rearranging for $\lambda$, we get $\lambda = \frac{v}{\nu}$. Substituting the given values, $\lambda = \frac{340}{1700} = 0.2 \, \text{m}$.

Key Concept: Multiple Reflection of Sound

b) Megaphone

[Solution Description]

Devices like megaphones, horns, and musical instruments such as trumpets are designed to send sound in a particular direction without spreading it in all directions. This is achieved by using the principle of multiple reflections of sound.

Key Concept: Effect of Temperature on Echo

b) 17.0 meters

[Solution Description]

The minimum distance is calculated using the formula:

$\text{Distance} = \frac{\text{Speed of Sound} \times \text{Time Interval}}{2}.$

Given the speed of sound is $340 \, \text{m/s}$ and the time interval is 0.1 seconds, the minimum distance is:

$\text{Distance} = \frac{340 \times 0.1}{2} = 17.0 \, \text{m}.$

Key Concept: Wavelength

c) Wavelength

[Solution Description]

The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength, denoted by $\lambda$. Wavelength is a key characteristic of sound waves and is defined as the spatial period of the wave—the distance over which the wave’s shape repeats.

Key Concept: Frequency and Wavelength

b) 0.68 meters

[Solution Description]

The relationship between the speed ($v$), frequency ($\nu$), and wavelength ($\lambda$) of a sound wave is given by the equation $v = \lambda \nu$. To find the wavelength, we rearrange this equation to solve for $\lambda$: $\lambda = \frac{v}{\nu}$. Substituting the given values, $\lambda = \frac{340}{500} = 0.68$ meters. Therefore, the wavelength of the sound wave is 0.68 meters.

Key Concept: Uses of Multiple Reflection of Sound

c) To direct sound in a specific direction

[Solution Description]

Megaphones are designed to direct sound in a particular direction without spreading it in all directions. This is achieved by using multiple reflections of sound within the conical structure, which helps to amplify and focus the sound waves towards the intended audience.

Key Concept: Sound Waves are Longitudinal Waves

c) Longitudinal wave

[Solution Description]

Sound waves propagate through a medium as a series of compressions and rarefactions. The particles of the medium oscillate back and forth parallel to the direction of wave propagation, which is characteristic of longitudinal waves.