Правка разделов «Statement», «Solution»

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		@@ -1,11 +1,9 @@
			### Statement

		−	$3.2.15.$ [Insert the problem statement]
		+	$3.2.15.$ The elevator rises and descends in a vertical shaft 400 m deep, completing the trip in 40 s. It first accelerates with constant acceleration, and then decelerates with an acceleration equal in magnitude to the initial one. By how much will the pendulum clock in the elevator lag behind a stationary clock over the course of a day? The elevator is in motion for 5 hours each day.

			### Solution

		−
		−	#### Understanding the motion


			The elevator descends a distance $S=400\,\mathrm{m}$ in $t=40\,\mathrm{s}$.
			During the first half of the descent it accelerates with acceleration $a$,
			and during the second half it decelerates with the same magnitude.

			The total distance traveled is

			\[
			S=\frac{1}{2}a\left(\frac{t}{2}\right)^2+
			\frac{1}{2}a\left(\frac{t}{2}\right)^2
			=\frac{at^2}{4}.
			\]

			Therefore,

			\[
			a=\frac{4S}{t^2}
			=\frac{4\cdot400}{40^2}
			=1\,\mathrm{m/s^2}.
			\]

			During the first half of the descent, the effective gravitational acceleration
			for the pendulum is

			\[
			g_{\mathrm{eff},1}=g-a,
			\]

			so its period is

			\[
			T_1=2\pi\sqrt{\frac{l}{g-a}}.
			\]

			During the second half,

			\[
			g_{\mathrm{eff},2}=g+a,
			\]

			and therefore

			\[
			T_2=2\pi\sqrt{\frac{l}{g+a}}.
			\]

			Hence, during the first and second halves of the descent, the numbers of
			oscillations are

			\[
			n_1=\frac{t/2}{T_1}
			=\frac{t}{4\pi}\sqrt{\frac{g-a}{l}},
			\]

			and

			\[
			n_2=\frac{t/2}{T_2}
			=\frac{t}{4\pi}\sqrt{\frac{g+a}{l}}.
			\]

			Thus the total number of oscillations during one descent is

			\[
			n_{\mathrm{d}}
			=\frac{t}{4\pi\sqrt{l}}
			\left(\sqrt{g-a}+\sqrt{g+a}\right).
			\]

			If the elevator were not accelerating, the pendulum would make

			\[
			n_0=\frac{t}{2\pi}\sqrt{\frac{g}{l}}
			\]

			oscillations in the same time.

			The number of oscillations lost is therefore

			\[
			n_0-n_{\mathrm{d}}
			=\frac{t\sqrt{g}}{4\pi\sqrt{l}}
			\left[
			2-\sqrt{1-\frac{a}{g}}
			-\sqrt{1+\frac{a}{g}}
			\right].
			\]

			The normal period of the pendulum is

			\[
			T_0=2\pi\sqrt{\frac{l}{g}}.
			\]

			Therefore, the time lost by the clock during one descent is

			\[
			\Delta t=(n_0-n_{\mathrm{d}})T_0.
			\]

			After cancellation,

			\[
			\boxed{
			\Delta t=
			\frac{t}{2}
			\left[
			2-\sqrt{1-\frac{a}{g}}
			-\sqrt{1+\frac{a}{g}}
			\right]
			}.
			\]

			For $t=40\,\mathrm{s}$, $a=1\,\mathrm{m/s^2}$, and
			$g=9.8\,\mathrm{m/s^2}$,

			$$
			\Delta t = 20
			\left[
			2-\sqrt{1-\frac{1}{9.8}}
			-\sqrt{1+\frac{1}{9.8}}
			\right]
			\approx 0.054\,\mathrm{s}.
			$$

			The same time loss occurs during an ascent, since the two effective
			accelerations $g-a$ and $g+a$ simply occur in the opposite order.

			In $5$ hours,

			\[
			5\,\mathrm{h}=18000\,\mathrm{s},
			\]

			so the number of ascents or descents is

			\[
			N=\frac{18000}{40}=450.
			\]

			Consequently, the total time lost is

			\[
			\Delta T=N\Delta t
			=450(0.054)
			\approx24.3\,\mathrm{s}.
			\]

			Therefore,

			\[
			\boxed{\Delta T\approx24\,\mathrm{s}}.
			\]



			#### Answer
			\[
			\boxed{\Delta T\approx24\,\mathrm{s}}.
			\]
		ещё строк без изменений 158

	@@ -1,11 +1,9 @@
	### Statement		### Statement

	$3.2.15.$ [Insert the problem statement]		$3.2.15.$ The elevator rises and descends in a vertical shaft 400 m deep, completing the trip in 40 s. It first accelerates with constant acceleration, and then decelerates with an acceleration equal in magnitude to the initial one. By how much will the pendulum clock in the elevator lag behind a stationary clock over the course of a day? The elevator is in motion for 5 hours each day.

	### Solution		### Solution


	#### Understanding the motion


	The elevator descends a distance $S=400\,\mathrm{m}$ in $t=40\,\mathrm{s}$.		The elevator descends a distance $S=400\,\mathrm{m}$ in $t=40\,\mathrm{s}$.
	During the first half of the descent it accelerates with acceleration $a$,		During the first half of the descent it accelerates with acceleration $a$,
	and during the second half it decelerates with the same magnitude.		and during the second half it decelerates with the same magnitude.

	The total distance traveled is		The total distance traveled is

	\[		\[
	S=\frac{1}{2}a\left(\frac{t}{2}\right)^2+		S=\frac{1}{2}a\left(\frac{t}{2}\right)^2+
	\frac{1}{2}a\left(\frac{t}{2}\right)^2		\frac{1}{2}a\left(\frac{t}{2}\right)^2
	=\frac{at^2}{4}.		=\frac{at^2}{4}.
	\]		\]

	Therefore,		Therefore,

	\[		\[
	a=\frac{4S}{t^2}		a=\frac{4S}{t^2}
	=\frac{4\cdot400}{40^2}		=\frac{4\cdot400}{40^2}
	=1\,\mathrm{m/s^2}.		=1\,\mathrm{m/s^2}.
	\]		\]

	During the first half of the descent, the effective gravitational acceleration		During the first half of the descent, the effective gravitational acceleration
	for the pendulum is		for the pendulum is

	\[		\[
	g_{\mathrm{eff},1}=g-a,		g_{\mathrm{eff},1}=g-a,
	\]		\]

	so its period is		so its period is

	\[		\[
	T_1=2\pi\sqrt{\frac{l}{g-a}}.		T_1=2\pi\sqrt{\frac{l}{g-a}}.
	\]		\]

	During the second half,		During the second half,

	\[		\[
	g_{\mathrm{eff},2}=g+a,		g_{\mathrm{eff},2}=g+a,
	\]		\]

	and therefore		and therefore

	\[		\[
	T_2=2\pi\sqrt{\frac{l}{g+a}}.		T_2=2\pi\sqrt{\frac{l}{g+a}}.
	\]		\]

	Hence, during the first and second halves of the descent, the numbers of		Hence, during the first and second halves of the descent, the numbers of
	oscillations are		oscillations are

	\[		\[
	n_1=\frac{t/2}{T_1}		n_1=\frac{t/2}{T_1}
	=\frac{t}{4\pi}\sqrt{\frac{g-a}{l}},		=\frac{t}{4\pi}\sqrt{\frac{g-a}{l}},
	\]		\]

	and		and

	\[		\[
	n_2=\frac{t/2}{T_2}		n_2=\frac{t/2}{T_2}
	=\frac{t}{4\pi}\sqrt{\frac{g+a}{l}}.		=\frac{t}{4\pi}\sqrt{\frac{g+a}{l}}.
	\]		\]

	Thus the total number of oscillations during one descent is		Thus the total number of oscillations during one descent is

	\[		\[
	n_{\mathrm{d}}		n_{\mathrm{d}}
	=\frac{t}{4\pi\sqrt{l}}		=\frac{t}{4\pi\sqrt{l}}
	\left(\sqrt{g-a}+\sqrt{g+a}\right).		\left(\sqrt{g-a}+\sqrt{g+a}\right).
	\]		\]

	If the elevator were not accelerating, the pendulum would make		If the elevator were not accelerating, the pendulum would make

	\[		\[
	n_0=\frac{t}{2\pi}\sqrt{\frac{g}{l}}		n_0=\frac{t}{2\pi}\sqrt{\frac{g}{l}}
	\]		\]

	oscillations in the same time.		oscillations in the same time.

	The number of oscillations lost is therefore		The number of oscillations lost is therefore

	\[		\[
	n_0-n_{\mathrm{d}}		n_0-n_{\mathrm{d}}
	=\frac{t\sqrt{g}}{4\pi\sqrt{l}}		=\frac{t\sqrt{g}}{4\pi\sqrt{l}}
	\left[		\left[
	2-\sqrt{1-\frac{a}{g}}		2-\sqrt{1-\frac{a}{g}}
	-\sqrt{1+\frac{a}{g}}		-\sqrt{1+\frac{a}{g}}
	\right].		\right].
	\]		\]

	The normal period of the pendulum is		The normal period of the pendulum is

	\[		\[
	T_0=2\pi\sqrt{\frac{l}{g}}.		T_0=2\pi\sqrt{\frac{l}{g}}.
	\]		\]

	Therefore, the time lost by the clock during one descent is		Therefore, the time lost by the clock during one descent is

	\[		\[
	\Delta t=(n_0-n_{\mathrm{d}})T_0.		\Delta t=(n_0-n_{\mathrm{d}})T_0.
	\]		\]

	After cancellation,		After cancellation,

	\[		\[
	\boxed{		\boxed{
	\Delta t=		\Delta t=
	\frac{t}{2}		\frac{t}{2}
	\left[		\left[
	2-\sqrt{1-\frac{a}{g}}		2-\sqrt{1-\frac{a}{g}}
	-\sqrt{1+\frac{a}{g}}		-\sqrt{1+\frac{a}{g}}
	\right]		\right]
	}.		}.
	\]		\]

	For $t=40\,\mathrm{s}$, $a=1\,\mathrm{m/s^2}$, and		For $t=40\,\mathrm{s}$, $a=1\,\mathrm{m/s^2}$, and
	$g=9.8\,\mathrm{m/s^2}$,		$g=9.8\,\mathrm{m/s^2}$,

	$$		$$
	\Delta t = 20		\Delta t = 20
	\left[		\left[
	2-\sqrt{1-\frac{1}{9.8}}		2-\sqrt{1-\frac{1}{9.8}}
	-\sqrt{1+\frac{1}{9.8}}		-\sqrt{1+\frac{1}{9.8}}
	\right]		\right]
	\approx 0.054\,\mathrm{s}.		\approx 0.054\,\mathrm{s}.
	$$		$$

	The same time loss occurs during an ascent, since the two effective		The same time loss occurs during an ascent, since the two effective
	accelerations $g-a$ and $g+a$ simply occur in the opposite order.		accelerations $g-a$ and $g+a$ simply occur in the opposite order.

	In $5$ hours,		In $5$ hours,

	\[		\[
	5\,\mathrm{h}=18000\,\mathrm{s},		5\,\mathrm{h}=18000\,\mathrm{s},
	\]		\]

	so the number of ascents or descents is		so the number of ascents or descents is

	\[		\[
	N=\frac{18000}{40}=450.		N=\frac{18000}{40}=450.
	\]		\]

	Consequently, the total time lost is		Consequently, the total time lost is

	\[		\[
	\Delta T=N\Delta t		\Delta T=N\Delta t
	=450(0.054)		=450(0.054)
	\approx24.3\,\mathrm{s}.		\approx24.3\,\mathrm{s}.
	\]		\]

	Therefore,		Therefore,

	\[		\[
	\boxed{\Delta T\approx24\,\mathrm{s}}.		\boxed{\Delta T\approx24\,\mathrm{s}}.
	\]		\]



	#### Answer		#### Answer
	\[		\[
	\boxed{\Delta T\approx24\,\mathrm{s}}.		\boxed{\Delta T\approx24\,\mathrm{s}}.
	\]		\]
	ещё строк без изменений 158