题目

Problem

A particle $P$ of mass $m$ is attached to one end of a light elastic spring of natural length $2l$ . The other end of the spring is attached to a fixed point $A$ . The particle $P$ hangs in equilibrium vertically below $A$ , at the point $E$ where $AE=6l$ . The particle $P$ is then raised a vertical distance $2l$ and released from rest.

Air resistance is modelled as being negligible.

(a) Show that $P$ moves with simple harmonic motion of period $T$ where

T=4\pi\sqrt{\frac lg}

(8)

(b) Find, in terms of $m$ , $l$ and $g$ , the kinetic energy of $P$ as it passes through $E$

(3)

(c) Find, in terms of $T$ , the exact time from the instant when $P$ is released to the instant when $P$ has moved a distance $3l$ .

(4)

题目中文翻译

质量为 $m$ 的质点 $P$ 连接于一根原长为 $2l$ 的弹性轻弹簧的一端，另一端连接于固定点 $A$ 。质点 $P$ 悬挂在 $A$ 正下方并平衡于 $E$ 点，其中 $AE=6l$ 。随后将质点 $P$ 竖直提高距离 $2l$ ，再由静止释放。

模型中忽略空气阻力。

(a) 证明 $P$ 做周期为 $T$ 的简谐运动，其中

T=4\pi\sqrt{\frac lg}

(8)

(b) 用 $m$ 、 $l$ 和 $g$ 表示 $P$ 经过 $E$ 时的动能。

(3)

(4)

解答

(a)

解法一

思路

展开

先由平衡伸长求弹性模量，再从平衡位置向下量位移列运动方程。化为简谐形式后求角频率及周期。

答题过程

展开

Let the spring’s modulus be $\lambda$ . At equilibrium its extension is $6l-2l=4l$ , so

mg=\frac{\lambda(4l)}{2l},\qquad \lambda=\frac12mg.

Let $x$ be the downward displacement from $E$ . The spring’s extension is then $4l+x$ . Resolving downwards,

m\ddot x=mg-\frac{(mg/2)(4l+x)}{2l}.

Hence

m\ddot x=-\frac{mg}{4l}x,\qquad \ddot x=-\frac{g}{4l}x.

Thus the motion is SHM about $E$ , with $\omega^2=g/(4l)$ . The particle is released from rest at displacement $-2l$ , so the amplitude is $2l$ . The minimum extension is $4l-2l=2l>0$ . The period is

T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{4l}{g}}=\boxed{4\pi\sqrt{\frac lg}}.

(b)

解法一

思路

展开

经过平衡位置 E 时速率最大，利用振幅和角频率求速率，再换算动能。

答题过程

展开

At $E$ the speed is

v_{\max}=(2l)\sqrt{\frac{g}{4l}}=\sqrt{gl}.

Therefore the kinetic energy is

\boxed{\frac12mv_{\max}^2=\frac12mgl}.

(c)

解法一

思路

展开

释放点在 E 上方 2l，首次走过 3l 后在 E 下方 l。全段仍向下运动，距离就是该段位移；用正确初相位求首次到达时间。

答题过程

展开

Take $t=0$ at release, and continue to measure $x$ downwards from $E$ . The particle starts from rest at $x=-2l$ , so

x=-2l\cos(\omega t).

After travelling directly downwards through $3l$ , its displacement from $E$ is

x=-2l+3l=l.

It has not yet reached the lower turning point $x=2l$ , so no reversal has occurred. Therefore

l=-2l\cos(\omega t),\qquad \cos(\omega t)=-\frac12.

For the first arrival, $0<\omega t<\pi$ , giving $\omega t=2\pi/3$ . Since $T=2\pi/\omega$ ,

\boxed{t=\frac{2\pi}{3\omega}=\frac T3}.

M3 2022 Jan UNUSED Q6

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