波长公式
是否知道为什么在道路警告中使用红色?原因是红色在所有颜色中具有最高的波长。那么现在让我们简要讨论一下什么是波长和波。无论是在电视上还是在现实生活中,每个人都见过海浪。此外,冲浪者喜欢骑在上面。然而,人们不知道的是,有一个方程可以用来量化波长。此外,我们将描述波长、波长公式、其感应和固定模型。
海浪
它是指由于风的作用而在外层水的运动。此外,空气原子和水粒子之间的摩擦运动使能量从微风转移到水中。此外,在科学中,波是能量的交换。
波长
波长是波峰或波谷之间的距离,特别是电磁波的点称为其波长。波峰是波的最高点,称为地壳,谷是波的最低点,称为波谷。同样,有不同的事物在比较波中移动,类似于水、字符串、空气(声波)、地球或地面,光也可以被视为波。此外,我们用希腊字母 lambda (λ) 来描述波的波长。频率越高,波长越短。而且,波的波长等于波的速度除以频率。该单位用于描述波长,以米为单位。
波长公式
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.
在每秒或赫兹的特定时间周期内通过一个点的波峰。 (每秒周期或赫兹)
示例问题
问题 1:波的波长为 3.40 m。如果它是以下每种类型的波,请计算波的频率。取声速为 343 m/s,光速为 3.00 × 10 8 m/s。
- 一个声波
- 一个光波
解决方案:
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
问题2:求频率为8.6 GHz = 8.60 × 10 9 Hz (G = (Giga) = 10 9 ) 的电磁波的波长,该波长在微波范围内。
解决方案:
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
问题 3:如果频率为 578 Hz,波长为 2.40 m,求未知流体介质中声波的速度。
解决方案:
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
问题 4:一个信号在完成一个周期所需的时间内传播了 75 英尺的距离。它的频率是多少?
解决方案:
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
问题 5:EM 波的最大峰值相隔 8 英寸。频率是多少,
- 兆赫
- 千兆赫
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
问题6:如果波峰和波谷之间的距离是9米,求波长。
解决方案:
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.
问题 7:如果频率为 1156Hz,波长为 4.8m,求声波在未知流体介质中的速度。
解决方案:
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