NCERT Solutions For Class 9 Science Chapter 12 Sound

Bats generate high-pitched ultrasonic shrill. These high-pitched shrill are reflected by objects such as preys and comes back to the bat’s ear. Thus, a bat can easily determine the distance of a prey to catch it.

Loudness depends on the amplitude of the sound. The louder a sound, the more energy it has. Loudness is proportional to the square of the amplitude of vibrations.

Quality is the characteristic of the sound which helps us to recognise a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

The vibration of the medium that moves along the direction of the wave is known as longitudinal wave. In the case of sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Therefore, a sound wave is called a longitudinal wave.

Vibration in the particles causes sound. When a body vibrates, it forces the nearby particles of the medium to vibrate. Hence, a disturbance is created in the medium, which travels in the form of waves. This disturbance is called sound.

To clean an object when it is put in a cleaning solution, the ultrasonic sound waves are passed through the solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

The reverberation is the persistence of sound due to repeated reflection. When the source produces sound, it starts moving in all the directions. It is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Thus, sound can be heard even after the source has stopped producing the sound.

To minimise reverberation, sound should be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, heavy curtains, and cushioned seats can be used to reduce reverberation.

(i) Sound board: It is used to send the sound towards audience in a big hall or auditorium. This works on the basis of laws of reflection of sound waves.

(ii) Stethoscope: It is also based on reflection of sound. In a stethoscope, the sound of a patient’s heartbeat reaches to the doctor’s ear by multiple reflection of sound.

Yes, the incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Moreover, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

The speed of sound is very less as compared to the speed of light. Hence, sound of thunder takes more time to reach the Earth as compared to light. Therefore, a flash is seen before than the thunder is heard.

The ultrasounds cannot pass through a defective metal block but they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. Thus, defects in metal blocks can be detected by using ultrasound.

\begin{array}{l}\text{Given, Time taken to hear to echo, t = 5 s}\\ \text{Distance of the object from the submarine, s = 3625 m}\\ \therefore \text{ Total distance travelled = 2 s}\\ \therefore \text{ Velocity, v = }\frac{2\text{ s}}{t}\text{ = }\frac{2\times \text{ 3625 m}}{5}\\ {\text{ = 1450 ms}}^{-1}\end{array}

An echo can be heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Therefore, echo cannot be heard on a summer day as the time interval between the original sound and the reflected sound decreases.

For this, take an electric bell and hang it inside an empty bell-jar fitted with a vacuum pump. Now, start removing the air from the bell-jar with the help of a vacuum pump. It can be observed that the sound of the ringing bell decreases. On further pumping off air, vacuum will be created inside the jar. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. This experiment shows that sound cannot travel through vacuum. Hence, sound requires a material medium for its propagation.

When a body vibrates, it creates a region of high pressure and low pressure in its surrounding. These regions of high pressure and low pressure are called compressions and rarefactions. As the body continues to move forward and backward, it produces a series of compressions and rarefactions.

Pinna collects different sounds produced in our surroundings and sends these sounds to the ear drum through ear canal. At the end of ear canal, there is a thin membrane called the eardrum or tympanic membrane. When the compression of sound waves strikes the eardrum, it is pushed inward. When the rarefaction of sound waves strikes the ear drum, it moves outward. Thus, eardrum vibrates. Middle ear consists of three bones – hammer, anvil and stirrup. These bones amplify the vibrations from the eardrum and transmit these vibrations to the inner ear. Inner ear consists of cochlea that converts vibrations or pressure variations into electrical signals. These electrical signals sent to brain via auditory nerve. Brain interprets them as sound.

It is a device that uses ultrasound propagation to navigate, communicate or detect underwater objects such as submarine, a sunken ship, an iceberg, etc. A SONAR apparatus consists of two parts:

1. A transmitter (for emitting ultrasonic waves).
2. A receiver (for detecting the reflected ultrasonic waves).

Both these parts are installed in a ship or a boat. The transmitter sends ultrasonic waves towards the ocean floor.
 

These waves when reflected back by an object or the ocean floor in form of ‘echo’ are detected by a detector.

Depth of an object can be calculated using time interval between generation of wave and reception of its echo and the speed of sound in water. This method is known as sonar.

\begin{array}{l}\text{Given,}\text{Frequency}\text{of}\text{sound}\text{=}\text{100}\text{Hz}\\ \text{Total}\text{time}\text{=}\text{1}\text{min}\text{=}\text{60}\text{s}\\ \text{On using the relation,}\\ \text{Frequency = }\frac{\text{Number}\text{of}\text{oscillations}}{\text{Total}\text{time}}\\ \text{Here,}\text{Number}\text{of}\text{oscillations}\text{=}\text{Frequency}\text{×}\text{Total}\text{time}\\ \text{=}\text{100 Hz}\text{×}\text{60}\text{ s\hspace{0.17em}}\text{=}\text{6000}\\ \text{Thus,}\text{the}\text{source}\text{vibrate}\text{6000}\text{times}\text{in}\text{a}\text{minute}\text{.}\end{array}

\begin{array}{l}\text{Consider the length of the aluminium rod = l}\text{.}\\ \text{Speed of sound wave in aluminium at 25 °C,}\\ v_1=6420{\text{ ms}}^{\text{-1}}\\ Hence,\text{ time taken by the sound wave to reach}\\ {\text{the other end, t}}_1=\frac{d}{v_1}=\frac{d}{6420}\mathrm{...}\text{ (i)}\\ \text{Speed of sound in air at 25 °C,}\\ v_2={\text{ 346 ms}}^{\text{-1}}\\ Hence,\text{ the time taken by sound wave to reach the}\\ {\text{other end, t}}_2=\frac{d}{v_2}=\frac{d}{\text{346}}\mathrm{...}\text{ (ii)}\\ \text{From (i) and (ii),}\\ \frac{{\text{t}}_{\text{2}}}{{\text{t}}_{\text{1}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55\text{.}\end{array}