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Sound - Range of Hearing and Ultrasound

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The audible range of hearing for an average human being is from 20 Hz20 \text{ Hz} to 20,000 Hz20,000 \text{ Hz} (or 20 kHz20 \text{ kHz}).

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Sounds with frequencies below 20 Hz20 \text{ Hz} are termed as infrasonic sound or infrasound. Examples include sounds produced by whales, elephants, and rhinoceroses (as low as 5 Hz5 \text{ Hz}).

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Sounds with frequencies higher than 20 kHz20 \text{ kHz} are called ultrasonic sound or ultrasound. Bats, dolphins, and porpoises produce ultrasound.

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Ultrasounds are high-frequency waves that can travel along well-defined paths even in the presence of obstacles. They are used extensively in industry and for medical purposes.

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SONAR (Sound Navigation and Ranging) is a device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects. It works on the principle of echo-ranging.

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Medical applications of ultrasound include Echocardiography (ECG) to image the heart and Ultrasonography for examining internal organs or monitoring fetal growth.

📐Formulae

v=νλv = \nu \lambda

2d=v×t2d = v \times t

d=v×t2d = \frac{v \times t}{2}

💡Examples

Problem 1:

A sonar device on a submarine sends out a signal and receives an echo 5 s5 \text{ s} later. Calculate the distance of the object from the submarine if the speed of sound in water is 1530 m/s1530 \text{ m/s}.

Solution:

Given: Time t=5 st = 5 \text{ s}, Speed of sound v=1530 m/sv = 1530 \text{ m/s}. Using the echo-ranging formula: 2d=v×t2d = v \times t 2d=1530×52d = 1530 \times 5 2d=76502d = 7650 d=76502=3825 md = \frac{7650}{2} = 3825 \text{ m}

Explanation:

The ultrasonic signal travels from the submarine to the object and back, covering a total distance of 2d2d. By dividing the total distance by 22, we find the actual distance of the object.

Problem 2:

A sound wave has a frequency of 2 kHz2 \text{ kHz} and a wavelength of 35 cm35 \text{ cm}. How long will it take to travel 1.5 km1.5 \text{ km}?

Solution:

Frequency ν=2 kHz=2000 Hz\nu = 2 \text{ kHz} = 2000 \text{ Hz}, Wavelength λ=35 cm=0.35 m\lambda = 35 \text{ cm} = 0.35 \text{ m}. Speed v=νλ=2000×0.35=700 m/sv = \nu \lambda = 2000 \times 0.35 = 700 \text{ m/s}. Distance s=1.5 km=1500 ms = 1.5 \text{ km} = 1500 \text{ m}. Time t=sv=1500700≈2.14 st = \frac{s}{v} = \frac{1500}{700} \approx 2.14 \text{ s}.

Explanation:

First, calculate the speed of the wave using the relationship between speed, frequency, and wavelength. Then, use the basic speed formula to find the time taken to cover the specified distance.