简单摆 - 定义、公式、推导、示例

通常将一个点质量固定在一根轻且不可伸长的字符串上并悬挂在恒定支撑上的装置称为单摆。单摆的平均位置通常是垂直线经过的点。摆的长度用 L 表示，它是从悬挂点到悬挂物体的质心的垂直距离，前提是它处于平均位置。这种类型的摆往往具有单一的共振频率。

基本上，一个简单的钟摆是一种展示周期性运动的机械装置。它有一个小圆形摆锤，由一根不可伸长的字符串从任何固定端悬挂，长度为 L。它执行振荡运动，由引力驱动并发生在垂直平面内。一般来说，字符串最末端的悬垂摆锤是无质量的。可以通过增加字符串的长度来增加钟摆的时间周期。摆的时间周期与悬吊的摆锤的质量无关。钟摆的时间周期通常取决于摆锤的位置和重力加速度，因为它在地球上的每个地方并不均匀。

简单的摆公式

我们可以看到钟摆用于各种事物，例如时钟，或者如果您敏锐地观察，甚至花园中的秋千也像钟摆一样。让我们进一步研究一下。重要公式，

时间周期 = 2π/ω ₀ = 2π × √(L/g)

势能 = mgL(1-cosθ)

动能 = (½)mv ²

总能量 = 动能 + 势能 = (½)mv ² + mgL(1 – cosθ) = 常数。

一些基本术语

• 摆动运动——摆锤在周期性运动中进行的任何往复运动，当摆锤处于其中心位置时，该位置称为平衡位置。
• 时间周期——通常是钟摆完成一次完整振荡所用的总时间，用“T”表示。
• 振幅——平衡位置和摆的极端位置之间的距离。
• 长度–字符串的长度通常是字符串的固定端到质心之间的距离。

单摆时间段的推导

Provided condition,

1. It is a frictionless surrounding.
2. The arms of the pendulum are stiff and massless.
3. Acceleration due to gravity is constant.
4. The motion of the pendulum is in an accurate plane.

We can use the motion equation,

T – mg cosθ = mv²L

Here, torque brings mass to the equilibrium position,

Τ = mgL × sinθ = mgsinθ × L = I × α

Sin θ ≈ θ, for small angles of oscillations.

So,  Iα = -mgLθ

Α = -(mgLθ)/I

-ω₀² θ = -(mgLθ)/I

ω₀² θ = (mgL)/I

ω₀ = √(mgL/I)

By using, I = ML²

Here, I = moment of inertia of the bob

ω₀ = √(g/L)

Hence, the time period of the simple pendulum is given by,

T = 2π/ω₀ = 2π × √(L/g)

单摆势能的推导

As we know the basic equation of potential energy is,

Potential energy = mgh

Here, m = mass of the object, g = acceleration due to gravity, and h = height of the object.

In simple pendulum movement, the pendulum is constrained by the string. The height here is written in terms of the angle and length of the string.

Therefore, height = L(1 – cosθ)

If θ = 90°, then the pendulum is considered to be at its highest point.

Cos90° = 0, and h = L, then potential energy = mgL

If θ = 0°, then the pendulum is considered to be at its lowest point.

Cos0° = 1, and h = L(1 – 1) = 0, then potential energy = 0

Therefore, potential energy at all points is given as,

Potential energy = mgL(1-cosθ)

总能量的推导

As we know the basic equation of kinetic energy is,

Kinetic energy = (½)mv²

Here, m is the mass of the pendulum and v is the velocity of the pendulum.

Kinetic Energy = 0, at maximum displacement, and is maximum at zero displacements. Though the total energy is a constant being a function of time.

The mechanical energy of the pendulum.

In a simple pendulum, the mechanical energy of a simple pendulum remains to be conserved.

Total Energy = Kinetic Energy + Potential Energy =  (½)mv²+ mgL(1-cosθ) = constant.

要记住的要点

• 系统温度的变化可能会影响钟摆的时间周期，因为时间周期取决于钟摆的长度。
• 一个简单的钟摆往往被放置在一个非惯性参考系中。
• 如果摆的平均位置发生变化，'g'的值将被'g_有效'代替，以确定时间段。

示例 – 电梯以加速度“a”向上移动，则 T = 2π × √(L/g _eff ) = 2π √[L/(g + a)]

示例问题

问题1：什么是单摆？

回答：

An idealized body comprising of a bob or particle suspended to one end of the string and the other end fixed to a rigid support. Pendulum when pulled from one side tends to move in to and fro periodic motion and swings in vertical plane due to gravitational pull. This motion is oscillatory and periodic and is so termed as Simple Harmonic motion.

问题2：在一个单摆中，有效长度是多少？

回答：

Effective length in a simple pendulum is the length of the string from rigid support to the center of mass of the pendulum. The Center of mass of the pendulum is generally the center point of the bob. Simply, effective length is the distance between the suspension point to the center of the bob of the pendulum.

问题3：单摆的时间周期是2.4秒。钟摆的长度是多少？ (g=10m/s ² )。

解决方案：

Given,

T = 2.4seconds

To find,

Length of pendulum =?

Formula for time period,

T = 2π/ω₀ = 2π × √(L/g)  

2.4 =  2π × √(L/10)

(2.4)² = (2π)² (L/10)

L = 1.46m.

问题4：单摆的时间周期是1.2秒。钟摆的长度是多少？ (g = 10m/s ² )。

解决方案：

Given,

T = 1.2seconds

To find,

Length of pendulum = ?

Formula for time period,

T = 2π/ω₀ = 2π × √(L/g)  

1.2 =  2π × √(L/10)

(1.2)² = (2π)² (L/10)

L = 0.36m.

问题 5：求周期为 3.6 秒的钟摆的长度，然后求其频率。

解决方案：

Given,

T = 3.6seconds

To find,

Length of pendulum =?

Frequency =?

Formula for time period,

T = 2π/ω₀ = 2π × √(L/g)  

3.6 =  2π × √(L/10)

(3.6)² = (2π)² (L/10)

L = 3.2m.

Frequency = 1/T = 1/(3.6) = 0.27

问题6：月球表面的重力加速度为1.8 m/s ² 。如果单摆在地球上的时间是 3.6 秒，它在月球上的时间是多少？

解决方案：

Given,

T = 3.6seconds

g = 1.8m/s²

To find,

Length of pendulum = ?

Formula for time period,

On earth,

T = 2π/ω₀ = 2π × √(3.6/10)  

3.6 =  2π × √(L/10)

(3.6)² = (2π)² (L/10)

L = 3.2m

On the moon,

T = 2π/ω₀ = 2π × √(3.2/1.8)  

T =  2π × √(3.2/1.8)

T =  2π × 1.3

T = 8.164seconds.

Therefore, the time period of the pendulum on the moon is 8.164 seconds.

问题7：摆的长度是2m，时间周期是多少？ (g = 10m/s ² )。

解决方案：

Given,

L = 2m

g=10m/s²

To find,

Time period = ?

Formula for time period,

T = 2π/ω₀ = 2π × √(L/g)

T = 2π × √(2/10)

T = 2.80seconds.