Edits to “Solution”, “Understanding the motion”, “Answer”
en/3.2.15.md
+11 −17
| ### Statement | |||
| $3.2.15.$ [Insert the problem statement] | |||
| @@ -5,20 +5,17 @@Statement | |||
| ### Solution | |||
| − | |||
| − | \[ | ||
| − | \textbf{Solution} | ||
| − | \] | ||
| + | #### 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 | |||
| \[ | |||
| − | S= | ||
| − | \frac{1}{2}a\left(\frac{t}{2}\right)^2+ | ||
| + | S=\frac{1}{2}a\left(\frac{t}{2}\right)^2+ | ||
| \frac{1}{2}a\left(\frac{t}{2}\right)^2 | |||
| =\frac{at^2}{4}. | |||
| \] | |||
| 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 | |||
| @@ -75,8 +72,7 @@Understanding the motion | |||
| \[ | |||
| n_{\mathrm{d}} | |||
| − | = | ||
| − | \frac{t}{4\pi\sqrt{l}} | ||
| + | =\frac{t}{4\pi\sqrt{l}} | ||
| \left(\sqrt{g-a}+\sqrt{g+a}\right). | |||
| \] | |||
| 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 | |||
| @@ -92,8 +88,7 @@Understanding the motion | |||
| \[ | |||
| n_0-n_{\mathrm{d}} | |||
| − | = | ||
| − | \frac{t\sqrt{g}}{4\pi\sqrt{l}} | ||
| + | =\frac{t\sqrt{g}}{4\pi\sqrt{l}} | ||
| \left[ | |||
| 2-\sqrt{1-\frac{a}{g}} | |||
| -\sqrt{1+\frac{a}{g}} | |||
| \right]. | |||
| \] | |||
| 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, | |||
| \[ | |||
| \boxed{ | |||
| \Delta t= | |||
| \frac{t}{2} | |||
| \left[ | |||
| 2-\sqrt{1-\frac{a}{g}} | |||
| -\sqrt{1+\frac{a}{g}} | |||
| \right] | |||
| }. | |||
| \] | |||
| @@ -128,16 +123,14 @@Understanding the motion | |||
| For \(t=40\,\mathrm{s}\), \(a=1\,\mathrm{m/s^2}\), and | |||
| \(g=9.8\,\mathrm{m/s^2}\), | |||
| − | \[ | ||
| − | \Delta t | ||
| − | = | ||
| − | 20 | ||
| + | $$ | ||
| + | \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, | |||
| \[ | |||
| \boxed{\Delta T\approx24\,\mathrm{s}}. | |||
| \] | |||
| @@ -171,5 +164,6 @@Understanding the motion | |||
| #### Answer | |||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \[ | ||
| + | \boxed{\Delta T\approx24\,\mathrm{s}}. | ||
| + | \] | ||
| ### Statement | ### Statement | ||
| $3.2.15.$ [Insert the problem statement] | $3.2.15.$ [Insert the problem statement] | ||
| @@ -5,20 +5,17 @@Statement | |||
| ### Solution | ### Solution | ||
| #### Understanding the motion | |||
| \[ | |||
| \textbf{Solution} | |||
| \] | |||
| 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 | ||
| \[ | \[ | ||
| S= | S=\frac{1}{2}a\left(\frac{t}{2}\right)^2+ | ||
| \frac{1}{2}a\left(\frac{t}{2}\right)^2+ | |||
| \frac{1}{2}a\left(\frac{t}{2}\right)^2 | \frac{1}{2}a\left(\frac{t}{2}\right)^2 | ||
| =\frac{at^2}{4}. | =\frac{at^2}{4}. | ||
| \] | \] | ||
| 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 | ||
| @@ -75,8 +72,7 @@Understanding the motion | |||
| \[ | \[ | ||
| n_{\mathrm{d}} | n_{\mathrm{d}} | ||
| = | =\frac{t}{4\pi\sqrt{l}} | ||
| \frac{t}{4\pi\sqrt{l}} | |||
| \left(\sqrt{g-a}+\sqrt{g+a}\right). | \left(\sqrt{g-a}+\sqrt{g+a}\right). | ||
| \] | \] | ||
| 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 | ||
| @@ -92,8 +88,7 @@Understanding the motion | |||
| \[ | \[ | ||
| n_0-n_{\mathrm{d}} | n_0-n_{\mathrm{d}} | ||
| = | =\frac{t\sqrt{g}}{4\pi\sqrt{l}} | ||
| \frac{t\sqrt{g}}{4\pi\sqrt{l}} | |||
| \left[ | \left[ | ||
| 2-\sqrt{1-\frac{a}{g}} | 2-\sqrt{1-\frac{a}{g}} | ||
| -\sqrt{1+\frac{a}{g}} | -\sqrt{1+\frac{a}{g}} | ||
| \right]. | \right]. | ||
| \] | \] | ||
| 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, | ||
| \[ | \[ | ||
| \boxed{ | \boxed{ | ||
| \Delta t= | \Delta t= | ||
| \frac{t}{2} | \frac{t}{2} | ||
| \left[ | \left[ | ||
| 2-\sqrt{1-\frac{a}{g}} | 2-\sqrt{1-\frac{a}{g}} | ||
| -\sqrt{1+\frac{a}{g}} | -\sqrt{1+\frac{a}{g}} | ||
| \right] | \right] | ||
| }. | }. | ||
| \] | \] | ||
| @@ -128,16 +123,14 @@Understanding the motion | |||
| 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 | \Delta t = 20 | ||
| = | |||
| 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, | ||
| \[ | \[ | ||
| \boxed{\Delta T\approx24\,\mathrm{s}}. | \boxed{\Delta T\approx24\,\mathrm{s}}. | ||
| \] | \] | ||
| @@ -171,5 +164,6 @@Understanding the motion | |||
| #### Answer | #### Answer | ||
| \[ | |||
| [Insert a concise answer or boxed result] | \boxed{\Delta T\approx24\,\mathrm{s}}. | ||
| \] | |||