题目
Problem
7.
A particle Q Q Q O O O
The maximum speed of Q Q Q 12 m s − 1 12\,\mathrm{m\,s^{-1}} 12 m s − 1
When Q Q Q 2 m 2\,\mathrm{m} 2 m O O O Q Q Q 4 5 m s − 1 4\sqrt5\,\mathrm{m\,s^{-1}} 4 5 m s − 1
(a) Find the amplitude of the motion.
(4)
The maximum magnitude of the acceleration of Q Q Q A m s − 2 A\,\mathrm{m\,s^{-2}} A m s − 2
(b) Find the value of A A A
(1)
(c) Find the length of time, in any one complete oscillation, for which the magnitude of the acceleration of Q Q Q 1 2 A \dfrac12A 2 1 A
(4)
题目中文翻译
质点 Q Q Q O O O
Q Q Q 12 m s − 1 12\,\mathrm{m\,s^{-1}} 12 m s − 1
Q Q Q O O O 2 m 2\,\mathrm{m} 2 m 4 5 m s − 1 4\sqrt5\,\mathrm{m\,s^{-1}} 4 5 m s − 1
(a) 求运动的振幅。
(4)
Q Q Q A m s − 2 A\,\mathrm{m\,s^{-2}} A m s − 2
(b) 求 A A A
(1)
(c) 求任意一次完整振动中，Q Q Q 1 2 A \dfrac12A 2 1 A
(4)
解答
(a)
解法一
思路
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最大速率为 a ω a\omega aω a 2 ω 2 a^2\omega^2 a 2 ω 2
答题过程
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Let the amplitude be a m a\ \mathrm m a m ω \omega ω
a ω = 12 , a 2 ω 2 = 144. a\omega=12,\qquad a^2\omega^2=144. aω = 12 , a 2 ω 2 = 144. Using v 2 = ω 2 ( a 2 − x 2 ) v^2=\omega^2(a^2-x^2) v 2 = ω 2 ( a 2 − x 2 ) ∣ x ∣ = 2 |x|=2 ∣ x ∣ = 2
( 4 5 ) 2 = ω 2 ( a 2 − 4 ) , 80 = a 2 ω 2 − 4 ω 2 . (4\sqrt5)^2=\omega^2(a^2-4),
\qquad 80=a^2\omega^2-4\omega^2. ( 4 5 ) 2 = ω 2 ( a 2 − 4 ) , 80 = a 2 ω 2 − 4 ω 2 . Subtracting from 144 = a 2 ω 2 144=a^2\omega^2 144 = a 2 ω 2
64 = 4 ω 2 , ω = 4 s − 1 . 64=4\omega^2,\qquad\omega=4\ \mathrm{s^{-1}}. 64 = 4 ω 2 , ω = 4 s − 1 . Therefore
a = 12 4 = 3 m . \boxed{a=\frac{12}{4}=3\ \mathrm m.} a = 4 12 = 3 m .
(b)
解法一
思路
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加速度大小 ω 2 ∣ x ∣ \omega^2|x| ω 2 ∣ x ∣
答题过程
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The maximum acceleration magnitude is
A = a ω 2 = 3 ( 4 ) 2 = 48. A=a\omega^2=3(4)^2=48. A = a ω 2 = 3 ( 4 ) 2 = 48. Hence
A = 48 . \boxed{A=48}. A = 48 .
(c)
解法一
思路
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加速度小于最大值的一半，等价于位移大小小于振幅的一半。一个周期内有两次经过这个中心区间，需要累计四段中心到边界的时间。
答题过程
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Since ∣ x ¨ ∣ = ω 2 ∣ x ∣ |\ddot x|=\omega^2|x| ∣ x ¨ ∣ = ω 2 ∣ x ∣ A = ω 2 a A=\omega^2a A = ω 2 a
∣ x ¨ ∣ < A 2 ⟺ ∣ x ∣ < a 2 . |\ddot x|<\frac A2\quad\Longleftrightarrow\quad |x|<\frac a2. ∣ x ¨ ∣ < 2 A ⟺ ∣ x ∣ < 2 a . Reset t = 0 t=0 t = 0 O O O
x = a sin 4 t . x=a\sin4t. x = a sin 4 t . The first arrival at x = a / 2 x=a/2 x = a /2
sin 4 t = 1 2 , 4 t = π 6 , t = π 24 . \sin4t=\frac12,\qquad 4t=\frac\pi6,\qquad t=\frac\pi{24}. sin 4 t = 2 1 , 4 t = 6 π , t = 24 π . By symmetry there are four such time intervals in a complete oscillation. Excluding the isolated equality instants does not change their total duration. Hence the required time is
4 ( π 24 ) = π 6 s ≈ 0.524 s . \boxed{4\left(\frac\pi{24}\right)=\frac\pi6\ \mathrm s\approx0.524\ \mathrm s.} 4 ( 24 π ) = 6 π s ≈ 0.524 s . 