题目
Problem
3.
A particle P P P O O O
When P P P 0.02 m 0.02\,\mathrm{m} 0.02 m O O O P P P 0.3 m s − 1 0.3\,\mathrm{m\,s^{-1}} 0.3 m s − 1 P P P 0.5 m s − 2 0.5\,\mathrm{m\,s^{-2}} 0.5 m s − 2
(a) Find the period of the motion.
(4)
The amplitude of the motion is a a a
Find
(b) the value of a a a
(3)
(c) the total length of time during each complete oscillation for which P P P 1 2 a \dfrac12a 2 1 a O O O
(4)
题目中文翻译
质点 P P P O O O
P P P O O O 0.02 m 0.02\,\mathrm{m} 0.02 m 0.3 m s − 1 0.3\,\mathrm{m\,s^{-1}} 0.3 m s − 1 0.5 m s − 2 0.5\,\mathrm{m\,s^{-2}} 0.5 m s − 2
(a) 求运动的周期。
(4)
运动振幅为 a a a
求
(b) a a a
(3)
(c) 每次完整振动中，P P P O O O 1 2 a \dfrac12a 2 1 a
(4)
解答
(a)
解法一
思路
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简谐运动的加速度大小是 ω 2 ∣ x ∣ \omega^2|x| ω 2 ∣ x ∣
答题过程
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For SHM, x ¨ = − ω 2 x \ddot x=-\omega^2x x ¨ = − ω 2 x
0.5 = ω 2 ( 0.02 ) ⟹ ω 2 = 25 , ω = 5 s − 1 . 0.5=\omega^2(0.02)
\quad\Longrightarrow\quad \omega^2=25,\quad \omega=5\ \mathrm{s^{-1}}. 0.5 = ω 2 ( 0.02 ) ⟹ ω 2 = 25 , ω = 5 s − 1 . The period is therefore
T = 2 π ω = 2 π 5 s . \boxed{\mathcal T=\frac{2\pi}{\omega}=\frac{2\pi}{5}\ \mathrm s.} T = ω 2 π = 5 2 π s .
(b)
解法一
思路
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将同一位置处的速率代入 v 2 = ω 2 ( a 2 − x 2 ) v^2=\omega^2(a^2-x^2) v 2 = ω 2 ( a 2 − x 2 )
答题过程
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For SHM,
v 2 = ω 2 ( a 2 − x 2 ) . v^2=\omega^2(a^2-x^2). v 2 = ω 2 ( a 2 − x 2 ) . Thus
0.3 2 = 25 ( a 2 − 0.02 2 ) , a 2 = 0.09 25 + 0.0004 = 0.004. 0.3^2=25(a^2-0.02^2),
\qquad a^2=\frac{0.09}{25}+0.0004=0.004. 0. 3 2 = 25 ( a 2 − 0.0 2 2 ) , a 2 = 25 0.09 + 0.0004 = 0.004. Since the amplitude is positive,
a = 10 50 m ≈ 0.0632 m . \boxed{a=\frac{\sqrt{10}}{50}\ \mathrm m\approx0.0632\ \mathrm m.} a = 50 10 m ≈ 0.0632 m .
(c)
解法一
思路
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所求是每个完整周期内的累计时间。将计时起点设在中心，求从中心到半振幅的时间；一个周期内有四段这样的时间。
答题过程
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Reset time to τ = 0 \tau=0 τ = 0 P P P O O O
x = a sin 5 τ . x=a\sin5\tau. x = a sin 5 τ . The first arrival at x = a / 2 x=a/2 x = a /2
sin 5 τ = 1 2 , 5 τ = π 6 , τ = π 30 . \sin5\tau=\frac12,
\qquad 5\tau=\frac\pi6,
\qquad \tau=\frac\pi{30}. sin 5 τ = 2 1 , 5 τ = 6 π , τ = 30 π . In each complete oscillation there are two passages through the central region − a / 2 ≤ x ≤ a / 2 -a/2\le x\le a/2 − a /2 ≤ x ≤ a /2
4 ( π 30 ) = 2 π 15 s ≈ 0.419 s . \boxed{4\left(\frac\pi{30}\right)=\frac{2\pi}{15}\ \mathrm s\approx0.419\ \mathrm s.} 4 ( 30 π ) = 15 2 π s ≈ 0.419 s . 