题目

Problem

A smooth bead of weight $12\,\mathrm{N}$ is threaded onto a light elastic string of natural length $3\,\mathrm{m}$ . The points $A$ and $B$ are on a horizontal ceiling, with $AB=3\,\mathrm{m}$ . One end of the string is attached to $A$ and the other end of the string is attached to $B$ .

The bead hangs freely in equilibrium, $2\,\mathrm{m}$ below the ceiling, as shown in Figure 2.

(a) Find the tension in the string.

(4)

(b) Show that the modulus of elasticity of the string is $11.25\,\mathrm{N}$ .

(2)

The bead is now pulled down to a point vertically below its equilibrium position and released from rest.

(2)

题目中文翻译

一个重为 $12\,\mathrm{N}$ 的光滑小珠穿在一根原长为 $3\,\mathrm{m}$ 的弹性轻绳上。点 $A$ 、 $B$ 位于水平天花板上， $AB=3\,\mathrm{m}$ 。绳的一端固定于 $A$ ，另一端固定于 $B$ 。

如图 2 所示，小珠自由悬挂并处于平衡，位于天花板下方 $2\,\mathrm{m}$ 处。

(a) 求绳中的张力。

(4)

(b) 证明绳的弹性模量为 $11.25\,\mathrm{N}$ 。

(2)

现在将小珠向下拉至其平衡位置正下方的一点，然后由静止释放。

(2)

解答

(a)

解法一

思路

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光滑小珠使两侧张力相同，水平分量平衡又使两侧倾角相同。由半跨 1.5 米及下垂 2 米求单段长度，再作竖直平衡。

答题过程

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The bead is smooth, so the tension $T$ is the same on both sides. Horizontal equilibrium requires equal inclinations, so the bead lies below the midpoint of $AB$ . Each stretched portion has length

\sqrt{1.5^2+2^2}=2.5\text{ m}.

Let each portion make angle $\theta$ with the vertical. Then

\cos\theta=\frac{2}{2.5}=\frac45.

Resolving vertically,

2T\cos\theta=12,\qquad 2T\frac45=12.

Therefore

\boxed{T=7.5\text{ N}}.

(b)

解法一

思路

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整条绳的伸长是两段现长之和减原长。使用整条绳的原长 3 米，不能只取一段的现长作伸长。

答题过程

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The total stretched length is $2(2.5)=5$ m, so the extension is

e=5-3=2\text{ m}.

By Hooke’s law,

T=\frac{\lambda e}{l},\qquad 7.5=\frac{\lambda(2)}{3}.

Hence

\boxed{\lambda=11.25\text{ N}}.

(c)

解法一

思路

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小珠沿中垂线运动，下降得越深，绳张力及其竖直分量都越大。经过平衡位置时合力从助运动变为阻运动，速率达到最大；此时用平衡位置的总伸长算弹性势能。

答题过程

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The bead moves on the vertical line through its equilibrium position. Below equilibrium the upward elastic resultant exceeds the weight; above equilibrium it is smaller than the weight. Thus, as the bead rises after release, its speed is greatest when it passes through the equilibrium position. At that instant the total extension is $2$ m. The elastic energy in the whole string is

E=\frac{\lambda e^2}{2l}=\frac{11.25(2)^2}{2(3)}=\boxed{7.5\text{ J}}.

M3 2020 Oct Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程