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(i) B) 90 db

(ii) D) 10 db

• Infra sonics are the sounds of frequency less than 20 Hz. 
• Ultra sonics are the sounds of frequency greater than 20 kHz. 
• Hence the ultra sonics have larger frequency.

The two quantities that vary periodically at a place in air as a sound wave travels through it are density and pressure of particles.

Uses of Ultrasonic sound are:

1. To clean delicate ornaments and the tiny parts of a watch. 

2. To observe internal organs of the body. 

3. To detect tumors in the brain. 

4. To detect faults in a metal. 

5. In RADAR systems 

6. To Kill certain microbes and insects. 

7. SONAR (Sound navigation and ranging) is used to locate the seabed or the position of a ship.

(i)IV

(ii)I

(iii)II

(i)b, since amplitude is largest

(ii)a, since frequency is lowest

By reducing the amplitude of the sound wave, its loudness decreases.

Hint: Loudness of sound is proportional to the square of the amplitude.

Musical note is pleasant, smooth and agreeable to the ear while noise is harsh, discordant and displeasing to the ear. In musical note, waveform is regular while in noise waveform is irregular.

The frying pan will vibrate. We will not be able to hear the sound of vibration because sound cannot travel in vacuum.

1. The term infrasonic refer to sound waves below the frequencies of audible sound i.e. under 20 Hz.

2. Humans cannot hear infrasonic sound.

3. Movement of hands and falling of leaves comes below this audible range. Therefore we cannot hear these sounds.

(a) The distance covered by any particle of the medium when it completes one oscillation is equal to the wave length.

(b) Sound waves travelling in air are longitudinal waves.

(c) Wave length produced in a long slinky is longitudinal waves.

(d) Examples of transverse waves are:

1. All electromagnetic waves i.e. light waves, radio waves etc. are a transverse wave.

2. The water waves produced on the surface of the water are transverse waves. In water waves, the molecules of water move up and down from their mean position.

Examples of longitudinal waves:

1. Sound waves.

2. Waves produced in a slinky, by compressing a small portion of it and then releasing life.

(e) Frequency is the physical quantity, whose S.I. the unit is Hertz.

The answer is (C) 200 Hz

Loudness:

i. Loudness depends upon the intensity of vibration.

ii. Intensity of a wave is proportional to square of the amplitude (I ∝ A²) and is measured in the (SI) unit of W/m²

iii. The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud.

iv. Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is I₀ = 10^-12 W/m².

v. Loudness of a sound of intensity I

(measured in unit bel) is given by,

L₂ = log₁₀ \((\frac{I}{I_0})\) ………….. (1)

vi. Decibel is the commonly used unit for loudness.

vii. As, 1 decibel or 1 dB = 0.1 bel.

∴ 1 bel = 10 dB. 

Thus, loudness in dB is 10 times loudness in bel.

∴ L_dB = 10L_bel = 10 log₁₀ \((\frac{I}{I_0})\)

For sound of least audible intensity I₀

L_dB = 10 log₁₀ \((\frac{I_0}{I_0})\) = 10 log₁₀ (1) = 0 ………… (2)

This corresponds to threshold of hearing.

viii. For sound of 10 dB,

10 = 10 log₁₀ \((\frac{I}{I_0})\)

∴ \((\frac{I}{I_0})\) = 101 or I = 10 I0

For sound of 20 dB,

20 = 10 log₁₀ \((\frac{I}{I_0})\)

= \((\frac{I}{I_0})\) = 10² or I = 100 I₀ and so on.

ix. This implies, loudness of 20 dB sound is felt double that of 10 dB, but its intensity is 10 times that of the 10 dB sound. Similarly, 40 dB sound is left twice as loud as 20 dB sound but its intensity is 100 times as that of 20 dB sound and 10000 times that of 10 dB sound. This is the power of logarithmic or exponential scale. If we move away from a (practically) point source, the intensity of its sound varies inversely with square of the distance, i.e., I ∝ \(\frac{1}{r^2}.\)

A wave in which the vibrations of all the particles along the path of a wave are constrained to a single plane is called a polarised wave.

Given: λ_a = 1 m, λ_l = 2.5 m, v_a = 330 m/s,

To find: Speed of sound (v_l)

Formula: v = nλ

Calculation:

From formula,

Since the frequency n remains the same,

v_a = nλ_a and v_l = nλ_l

∴ \(\frac{v_l}{v_a}=\frac{λ_l}{λ_a}\)

∴ v_l = v_a \(\frac{λ_l}{λ_a}\) = 330 × \(\frac{2.5}{1}\) = 825 m/s

A loud sound is harsh to ears. Such sounds produce noise.

Noise pollution is harmful to humans in many ways. Constant exposure to noise pollution can create many health related problems; like insomnia, hypertension and may even lead to loss of hearing.

Some sources of noise pollution are as follows: 

→ Televisions and transistors running at high volumes 

→ Loudspeakers and crackers 

→ Horns of buses, cars and trucks 

→ Home appliances such as mixer, desert cooler, etc.

Correct option is C) quality

(a) (i) Diagram is showing the principal note.

(b) (iii)Diagram has frequency four times that of the first.

(c) Ratio is 1:2

(a) Loudness

(b) Quality

(c) Pitch.

(i) Loudness

(ii) Quality or timbre

(iii) Pitch

(a) Frequency 

(b) Amplitude 

(c) Waveform

(i) Frequency

(ii) Amplitude

(iii) Waveform

It is because the vibrations produced by the vocal chord of each person have a characteristic waveform which is different for different persons.

A stringed instrument is provided with the provision for adjusting the tension of the string. By varying the tension, we can get desired frequency.

No, They have same waveform.

Different instruments emit different subsidiary notes. A note played on one instrument has a large number of subsidiary notes while the same note when played on other instrument contains only few subsidiary notes. So they have different waveforms.

Waveforms

Explanation: The waveform of a sound depends on the number of the subsidiary notes and their relative amplitude along with the principal note. The resultant vibration obtained by the superposition of all these vibrations gives the waveform of sound.

Correct option is D) ma

Phenomenon is resonance 

(b) The frequency of the loud sound is increased compared to frequency of tuning fork. 

(c) Unit of loudness is decibel dB which signifies sound PRESSURE LEVEL. 

ldB = 10 log¹⁰ 1/10 

B) reflection from various angles

D) The disturbance travels

1. He is trying to match the frequency of the radio components with the brodcasting station he wants to receive and hence to produce resonance (make the sound louder). 

2. Reasonance is the phenomenon. 

3. Resonance “is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as Resonance.” 

1. Sonar is based on echo. 

2. The sensation of sound persists in our ears for about 1/10th, of a second after exciting stimulus ceases to act. If d is the distance between the observer and obstancle and v the speed of sound, the time taken to hear the echo is 

t = Distance travelled in going and coming back/speed of Sound of v = 340 ms^-1 

∴ Minimum distance of 17.5m from the listner, to hear. echo distinctly. 

Characteristic having 

1. Two sounds with equal loudness but having different frequencies is Pitch. 

2. PITCH : “is that characteric of sound by which an acute (or shrill) not can be distinguished from a grave or flat note.” 

3. Characteristic is quality.

Pitch of sound is determined by its wavelength or the frequency. Two notes of the same amplitude and sounded on the same instrument will differ in pitch when their vibrations are of different wavelengths or frequencies.

The ultrasonic waves are used for the sound ranging. Ultrasonic waves have a frequency more than 20,000 Hz but the range of audibility of human ear is 20Hz to 20,000 Hz

FREQUENCY i.e. number of vibrations per second determines the pitch. Higher frequency, higher pitch means shrill sound. A ‘ low pitch has flat sound.

Sonar is sound navigation and ranging. Ultrasonic waves are sent in all directions from the ship and they are received on their return after reflection from the obstacles. They use the method of echo.

• Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves. 
• The high pitched ultrasonic squeaks of the bat are reflected from the obstacles or prey and returned to bat’s ear. 
• The nature of reflections tells the bat where the obstacle or prey is and what it is like. 
• The bats use ultrasound for navigation and location of the food in dark.

distance (s) = 1360 m, 

time for first echo = 3 s, 

time for second echo = 5 s 

To Find : speed of sound (v)

Formula : speed = distance/time

Calculation :

Time for first echo = 3 s 

∴ time taken by sound to travel given distance t₁ 

= 3/2  = 1.5 s

Time for second echo = 5 s 

∴ time taken by sound to travel given distance t₂ 

= 5/2 = 2.5 s

∴Total time taken by sound to travel given distance, 

T = 1.5 + 2.5 = 4 s

From formula,

v = 1360/4

∴v = 340 m/s

(i) army : Echoes are used by army to locate the gun position of enemy. Radar an instrument is used to locate an enemy air-craft ship. 

(ii) geologists : Echoes are used by geologists for mineral prospecting.

(iii) fishermen : for locating fishes ultrasonic waves are sent into water. If these vibrations strike a fish, they are reflected back to the receivers. The time for hearing the echoes recorded. The position of fish is calculated by d = v × t/2 using vel. of sound in water as 1450ms^-1.

In order to determine the speed of sound in air, sound produced from a place at known distance d at least 50m from the reflecting surface. The time interval t in which the echo reaches the place from where sound was produced is noted by a stop watch. Then speed of sound is calculated as v = total distance travelled /time interval = 2d/t ms^-1