| \textbf{Solution} | | \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 |
| | | |
| \[ | | \[ |
| n_{\mathrm{d}} | | n_{\mathrm{d}} |
| = | | = |
| \frac{t}{4\pi\sqrt{l}} | | \frac{t}{4\pi\sqrt{l}} |
| \left(\sqrt{g-a}+\sqrt{g+a}\right). | | \left(\sqrt{g-a}+\sqrt{g+a}\right). |
| \] | | \] |
| | | |
| 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 |
| \(g=9.8\,\mathrm{m/s^2}\), | | \(g=9.8\,\mathrm{m/s^2}\), |
| | | |
| \[ | | \[ |
| \Delta t | | \Delta t |
| = | | = |
| 20 | | 20 |
| \left[ | | \left[ |
| 2-\sqrt{1-\frac{1}{9.8}} | | 2-\sqrt{1-\frac{1}{9.8}} |
| -\sqrt{1+\frac{1}{9.8}} | | -\sqrt{1+\frac{1}{9.8}} |
| \right] | | \right] |
| \approx 0.054\,\mathrm{s}. | | \approx 0.054\,\mathrm{s}. |
| \] | | \] |
| | | |
| 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, |
| | | |
