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A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s⁻¹ find its time period.
The frequency of the wave is calculated by dividing speed by wavelength: 344 meters per second divided by 3.44 meters equals 100 Hz. Since the time period is the inverse of frequency, dividing 1 by 100 Hz gives 0.01 seconds. Thus, the time taken for one complete density oscillation at any fixed poinRead more
The frequency of the wave is calculated by dividing speed by wavelength: 344 meters per second divided by 3.44 meters equals 100 Hz. Since the time period is the inverse of frequency, dividing 1 by 100 Hz gives 0.01 seconds. Thus, the time taken for one complete density oscillation at any fixed point in the medium is 0.01 seconds.
For more NCERT Solutions of Class 9 Science Exploration Chapter 10 Sound Waves Characteristics and Applications Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-10/
See lessA ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s⁻¹ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
The sonar pulse completes a two-way round trip by traveling to the ocean floor and reflecting back to the ship in 5 seconds. Therefore, the one-way transit time to reach the wreckage is 2.5 seconds. Multiplying the speed of sound in seawater, 1525 meters per second, by 2.5 seconds reveals that the wRead more
The sonar pulse completes a two-way round trip by traveling to the ocean floor and reflecting back to the ship in 5 seconds. Therefore, the one-way transit time to reach the wreckage is 2.5 seconds. Multiplying the speed of sound in seawater, 1525 meters per second, by 2.5 seconds reveals that the wreckage is located at a depth of 3812.5 meters.
For more NCERT Solutions of Class 9 Science Exploration Chapter 10 Sound Waves Characteristics and Applications Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-10/
See lessA vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s⁻¹.
The ultrasonic wave travels forward to the obstacle and reflects back, covering a total distance of 2.4 meters. Using the relation time equals total distance divided by speed, dividing 2.4 meters by 345 meters per second yields approximately 0.007 seconds. Therefore, the sensor detects the returningRead more
The ultrasonic wave travels forward to the obstacle and reflects back, covering a total distance of 2.4 meters. Using the relation time equals total distance divided by speed, dividing 2.4 meters by 345 meters per second yields approximately 0.007 seconds. Therefore, the sensor detects the returning echo in roughly 6.96 milliseconds, triggering the warning beep.
For more NCERT Solutions of Class 9 Science Exploration Chapter 10 Sound Waves Characteristics and Applications Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-10/
See lessThe speed of sound in air is about 331 m s⁻¹ at 0 °C and nearly 344 m s⁻¹ at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
At 22 °C, dividing 1720 meters by 344 meters per second gives a transit time of 5.0 seconds. When the temperature drops to 0 °C, dividing 1720 meters by 331 meters per second yields approximately 5.196 seconds. Subtracting the two transit times shows that the sound of thunder takes roughly 0.20 secoRead more
At 22 °C, dividing 1720 meters by 344 meters per second gives a transit time of 5.0 seconds. When the temperature drops to 0 °C, dividing 1720 meters by 331 meters per second yields approximately 5.196 seconds. Subtracting the two transit times shows that the sound of thunder takes roughly 0.20 seconds of extra time to arrive.
For more NCERT Solutions of Class 9 Science Exploration Chapter 10 Sound Waves Characteristics and Applications Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-10/
See lessThe variation of density of medium for a sound wave propagating with a speed of 340 m s⁻¹ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
(i) Wavelength: The marked distance of 8 cm spans across two full compression-to-compression cycles, meaning two wavelengths equal 8 cm. Thus, the wavelength is 4 cm, which equals 0.04 m. (ii) Frequency: Using the wave relation where frequency equals speed divided by wavelength, dividing 340 m s⁻¹ bRead more
(i) Wavelength: The marked distance of 8 cm spans across two full compression-to-compression cycles, meaning two wavelengths equal 8 cm. Thus, the wavelength is 4 cm, which equals 0.04 m. (ii) Frequency: Using the wave relation where frequency equals speed divided by wavelength, dividing 340 m s⁻¹ by 0.04 m gives a frequency of 8500 Hz (or 8.5 kHz).
For more NCERT Solutions of Class 9 Science Exploration Chapter 10 Sound Waves Characteristics and Applications Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-10/
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