Is it known why red colour is used at warnings on road? The reason is that red has the highest wavelength among all colours. So now let's discuss what is wavelength and waves in brief. Everyone has seen waves whether on TV or in real life. Additionally, surfers like to ride on them. However, what individuals don't know is that there is an equation that we use to quantify wavelength. Additionally, we will portray the wavelength, wavelength formula, its induction, and settled models.
Wave
It refers to the movement on the outer layer of water because of the action of wind. Additionally, the friction movement between the air atoms and the water particles makes energy be moved from the breeze to water. Additionally, in science wave is the exchange of energy.
Wavelength
The wavelength is the distance between the peaks or valleys of the wave, particularly the points of the electromagnetic wave is known as its wavelength. A peak is the highest point of a wave and it is called the crust, while the valley is the lowest point of a wave and is called a trough. Likewise, there are different things that move in comparative waves, similar to the water, strings, air (sound waves), the earth or ground, and light additionally can be treated as a wave. In addition, we describe the wavelength of the wave by the Greek letter lambda (λ ). The higher the frequency, the shorter is the wavelength. Moreover, the wavelength of the wave is equivalent to the speed of the wave, divided by the frequency. The unit is used to describe wavelength in meters.
Wavelength Formula
Wavelength = wave velocity /frequency
λ = v/f
Where,
λ is the wavelength, the distance between the peaks in meters (m).
v is the velocity of the waves that are moving in a direction in meter per second (m/s)
f is the frequency.
The wave peaks that go through a point in certain time cycles per second or hertz. (cycles per second or Hz)
Sample Problems
Question 1: A wave has a wavelength of 3.40 m. Calculate the frequency of the wave if it is each of the following types of waves. Take the speed of sound as 343 m/s and the speed of light as 3.00 × 108 m/s.
- A sound wave
- A light wave
Solution:
1. Given λ = 3.40 m; vs = 343 m/s
Frequency of sound wave f = vs / λ = 343/3.4 = 100.9 Hz
2. λ = 3.40m;
Speed of light v = c = 3 ×108m/s
Frequency of light wave f = v/λ = c/λ
= 3 × 108/3.4
= 0.882 × 108 Hz
= 88 ×106 Hz = 88 MHz
Question 2: Find the wavelength of an electromagnetic wave with frequency 8.6 GHz = 8.60 times 109 Hz (G = (Giga) = 109), which is in the microwave range.
Solution:
Electromagnetic wave moves with speed 3 ×108m/s
As wavelength, λ= v/f (given f = 8.60 times 109 Hz)
λ= 3 × 108/8.60 × 109
= 0.349 × 10-1
= 0.0349 m
Question 3: Find the speed of a sound wave in an unknown fluid medium if a frequency of 578 Hz has a wavelength of 2.40 m.
Solution:
Speed of sound depends on bulk modules of media. Therefore wavelength changes in different media.
Given λ = 2.40m & f = 578 Hz
As v = λ × f = 2.40 × 578
v = 1387.2 m/s
Question 4: A signal travels a distance of 75 feet in the time it takes to complete one cycle. What is its frequency?
Solution:
We know that distance between peaks is wavelength .
Distance traveled by wave in the time to complete one cycle is wavelength.
λ = 75 feet = 75 × 0.3048 = 22.86m,
f = c/λ
f = 3 × 108/ 22.86
f = 13.123 MHz
Question 5: The maximum peaks of an EM wave are separated by a distance of 8 inches. What is the frequency in,
- MHz
- GHz
We know that distance between peaks is wavelength .
λ = 8 inches (1 meter = 39.37inches),
λ = 8/39.37 m = 0.2032m
f = c/λ = 3 × 108/ 0.2032,
f =1.476 × 109Hz
1. f(MHz) = 1.476 × 109/106
f = 1476 MHz
2. f(GHz) =1.476 × 109/109
f(GHz) = 1.476 GHz
Question 6: If the distance between crest and trough is 9 meters find the wavelength.
Solution:
As wavelength is defined as distance between two crusts or distance between two troughs.
But given distance between crust and trough
And distance between 2crusts = 2 × distance between crust and trough
wavelength = 2 × 9 = 18 λ.
Question 7: Find the speed of a sound wave in an unknown fluid medium if a frequency of 1156Hz has a wavelength of 4.8 m.
Solution:
Speed of sound depends on bulk modules of media. Therefore wavelength changes in different media.
Given λ = 2.40m & f = 578 Hz
As v = λ × f = 2.40 × 578
v = 5548.8 m/s
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