幅度公式

变量与其平均值的最大偏差称为幅度。它是从粒子的平均位置绕平均位置来回运动的最大位移。周期性的压力变化、周期性的电流或电压变化、电场或磁场的周期性变化等等都具有幅度。

幅度没有特定的公式。它可以通过这些变化的方程或图形表示来获得。

什么是振幅？

波的最高位移称为振幅。此外，您将在本课程中了解幅度、幅度公式、公式推导和求解示例。此外，完成该主题后，您将能够理解幅度。

Amplitude refers to the greatest deviation from equilibrium that an item in periodic motion might display. A pendulum, for example, swings past its equilibrium point (straight down) before reaching its maximum distance from the centre.

此外，振幅的距离为 A。此外，摆的完整范围的大小为 2A。例如，波浪和弹簧遵循周期性运动。此外，由于正弦函数在 +1 和 -1 之间振荡，它可以用来描述周期性运动。

SI 单位：米是最显着的幅度单位 (m)。

振幅公式

The amplitude of a variable is the biggest variation from its mean value. The amplitude formula can be used to calculate the sine and cosine functions. Amplitude is represented by the letter A. The sine (or cosine) function has the following formula:

x = A sin (ωt + ϕ)   

or   

x = A cos (ωt + ϕ)

where,

• x = displacement of wave (meter)
• A = amplitude
• ω = angular frequency (rad/s)
• t = time period
• ϕ = phase angle

幅度公式也称为正弦或余弦函数的最大值和最小值的平均值。始终使用绝对幅度值。

示例问题

问题 1：考虑一个来回摆动的钟摆。此外，相移为 0 弧度。此外，摆为 14.0 cm 或 x = 0.140 m，时间为 t = 8.50 s。那么，振荡的幅度是多少？

解决方案：

Given that,

x = 0.140 m

ω = π radians/s

ϕ = 0

t = 8.50 s

So, we can find the value of amplitude by rearranging the formula:

x = A sin (ωt+ϕ) → A = xsin(ωt+ϕ)

A = xsin(ωt+ϕ)

So, A = 0.14msin[(πradians/s)(8.50s)+0]

A = 0.140msin(8.50π)

Moreover, the sine of 8.50 π can be solved (by keeping in mind that the values is in radians) with a calculator:

Sin(8.50 π) = 1

So, the amplitude at time t is 8.50s is:

A = 0.140msin(8.50π)

A = 0.140m1

A = 0.140 m

Therefore, the amplitude of the pendulum’s oscillation is A =0.140 m = 14.0 cm.

问题 2：假设弹簧向上和向下弹跳一个千斤顶玩具的头部。此外，振荡的角频率为 π/6 弧度/秒，相移 (φ) 为 0 弧度。弹跳也有 5.00 厘米的幅度。千斤顶相对于 6 秒内的平衡位置处于什么位置？

解决方案：

Since, as we know that:

x = A sin (ωt+ϕ)

x = (0.500 m) sin [(π/6radians/s)(6.00s) + 0]

x = (0.500 m) sin (π/6radians/s)

x = (0.500 m) (0.00)

x = 0.00 m

So, at time t =6.00 s, the head of the-jack-in-the-box is at position 0.00 m that is the equilibrium position.

问题 3：如果 y = 6 cos (7t + 1) 是一个波。找到它的幅度。

解决方案：

Given: equation of wave y = 6cos(7t + 1)

Using amplitude formula,

x= A cos (ωt + ϕ)

On comparing it with the wave equation:

A = 6

ω = 7

ϕ = 1

Therefore, the amplitude of the wave = 6 units. 

问题 4：波是 y = 2sin(4t)。找出它的幅度。

解决方案：

The wave equation y = 2sin(4t)

Using the formula for amplitude,

x = A sin(ωt + ϕ)

When comparing the wave equation to the equation of motion,

A = 2

ω = 4

ϕ = 0

As a result, the amplitude of the wave is 2 units.

问题 5：考虑一个盒子里的千斤顶玩具，它的头部在弹簧上上下弹跳。此外，振荡的角频率为=π/6弧度/秒，相移为φ= 0弧度。此外，弹跳幅度为 5.00 cm。那么，Jack-in-the-head 相对于 1 秒内的平衡位置处于什么位置。

回答：

x = A sin (ωt+ϕ)

x = (0.500 m) sin [(π/6radians/s)(1.00s) + 0]

x = (0.500 m) sin (π/6radians/s)

x = (0.500 m) (0.500)

x = 0.250 m

x = 2.50 cm

So, at time 1.00 s the head of the jack-in-the-box is 2.5 cm above the equilibrium position.