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Transcript
彈簧水平振和垂直振,週期會不會一樣? 15-2 THE BLOCK-SPRING SYSTEM We begin by considering the dynamics of a block that is oscillating at the end of a massless spring, as in Fig. 15-4. We assume that the net force acting on the block is that exerted by the spring, which is given by Hooke’s law: Fsp kx where x is the displacement from the equilibrium position. When x is positive, Fsp is positive, the force is directed to the right. Thus, the force always tends to restore the block to its equilibrium position x = 0. Newton’s second law (F = ma) applied to the block is –kx = ma, which means k a x m The acceleration is directly proportional to the displacement, but is in the opposite direction, as is required for SHM. Since a = d2t/dt2, we have d 2x k (15-6) x 0 dt 2 m The differential equation is merely another way of writing Newton’s second law. When Eq. 15-6 is compared with Eq. 15-5, we see that the block-spring system executes simple harmonic motion with an angular frequency k (15-7) m or a period 2 m (15-8) T 2 k As is required for SHM, the period is independent of the amplitude. For a given spring constant, the period increase with the mass of the block: A more massive block oscillates more slowly. For a given block, the period decreases as k increases: A stiffer spring produces quicker oscillations. 圖 Figure 15-4 A block oscillating at the end of a spring. The restoring force is proportional to the displacement from equilibrium. EXAMPLE 15-4: Show that a block handing from a vertical spring, as in Fig. 15-7, executes simple harmonic motion. Solution: The block is subject to two forces: the upward force exerted by the spring and the downward force due to gravity. If x0 is the equilibrium extension at which these two forces balance, then mg kx 0 For any extension x, the net force on the block is F mg kx k x x 0 kx ' where x’ = x – x0 is the displacement from the equilibrium position. Since the restoring force is linearly proportional to the displacement from equilibrium, the motion will be simple harmonic. 圖 Figure 15-7 A block oscillating on a vertical spring executes simple harmonic motion about the equilibrium position. Benson 15.4 阻尼振盪 主題:DAMPED OSCILLATIONS Thus far we have ignored the inevitable energy losses that occur in real situations. Such losses may arise from external fluid resistance or from “internal friction” within a system. The energy, and consequently the amplitude, of such a damped oscillator decrease in time. To formulate the equation for damped oscillations, we consider the situation depicted in Fig. 15-13 which shoes a block immersed in a liquid. When the velocity is low, the damping is due to a resistive force f that is proportional to the velocity (see Section 6.4) : f v (15-20) where γ, measured in kg/s, is the damping constant. If we ignore the buoyancy of the fluid, Newton’s second law applied to the block is dx d 2x F kx m 2 dt dt where x is the displacement from equilibrium. (The prime used in Example 15-4 is dropped for simplicity.). This equation may be written in the form d 2x dx (15-21) m 2 kx 0 dt dt 圖 Figure 15-13 The oscillations of a block are damped when it is immersed in a fluid. In a real system, energy losses within the spring itself lead to damping. This form of differential equation arises in other mechanical or nonmechanical damped oscillations. Experience tells us that the mass will oscillate with ever-decreasing amplitude. As you can verify by substitution, the solution to Eq. 15-21 is t x A0e 2m cos ' t The damped angular frequency ω’ is given by (15-22) ' (15-23) 2m The damped angular frequency ω’ is less than the natural angular frequency, 0 k m . For ω’ to be real, the condition γ/2m < ω0, or equivalently γ < 2mω0, must be satisfied. When ω’ is real, the oscillations are underdamped, as illustrated in Fig. 15-14. The amplitude decays according to 2 2 0 t At A0 e 2m (Underdamped) (15.24) The damped period is T ' 2 ' . 圖 Figure 15-14 In an underdamped oscillation the system oscillates with an exponentially decaying amplitude. When the damping is so large that γ > 2mω0, ω’ is an imaginary number. In this case there is no oscillation and the system moves slowly back to its equilibrium position, as shown in Fig. 15-15. Hinged doors that close automatically and cueing devices on tone arms of turntable are overdamped. 圖 Figure 15-15 In critical damping (γ = 2mω0) the system approaches the equilibrium position most rapidly. In an overdaamped system (γ > 2mω0), the system approaches equilibrium slowly. When γ = 2mω0, we have ω’ = 0 and again there is no oscillation. This condition of critical damping leads to the shortest time for the system to return to equilibrium and is also shown in Fig. 15-15. Critical damping is used in the movements of electrical meters to damp the oscillations of the needle. The suspension system of a car is adjusted to have somewhat less than critical damping. When a fender is pressed down and released, the car executes perhaps one and a half oscillations before coming to rest. EXAMPLE 15-15: A 0.5kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency. (a) What is the damping constant? (b) How does the amplitude vary in time? (c) What is the critical damping constant? Solution: (a) The natural angular frequency is 0 k m 5 rad s . The damped angular frequency is ω’ = 0.998, ω0 = 4.99 rad/s. From Eq. 15-23. 2 4m 2 02 ' 2 This yields γ = 0.316 kg/s. (b) From Eq. 15-24 At A0 e 0.346 t (c) The critical damping constant is 2m0 5 kg s This is considerably larger than that found in part (a). 主題:FORCED OSCILLATIONS AND RESONANCE The loss in energy of a damped oscillator may be compensated for by work done by an external agent. For example, a child on a swing can be kept in motion by appropriately timed pushes (Fig. 15-16). In many instances, the external driving force varies sinusoidally at some angular frequency ωe. We assume F(t) = F0cosωet. Newton’s second law applied to such a forced or driven oscillator yields d 2x dx m 2 kx F0 cos e t (15-25) dt dt 圖 Figure 15-16 A child can be kept swinging by appropriately timed pushes. When the force is first applied, the motion is complex. However, the system ultimately settles into a stead-state oscillation balanced by the external input. The steady-state solution to Eq. 12-25 is (15-26) x A cose t where δ is the phase angle between the displacement x and the external force F. Notice that the amplitude is constant in time and that ωe is the angular frequency of the external driving force. When Eq. 15-26 is substituted into Eq. 15-25, we are led to conclude (the details are omitted) F0 m (15-27) A e 2 2 02 e2 e m Each driving frequency is characterized by its own amplitude, as shown in Fig. 15-17. At ωe = 0, the amplitude is merely the static extension F0 m 02 F0 k . As the external angular frequency ωe is increased, the amplitude rises until it reaches a maximum at ωmax, which is somewhat below ω0. At higher frequencies, the amplitude again decreases. Such a response is called resonance and ωmax is called the resonance angular frequency. When γ is small, the resonance curve is narrow and the peak occurs close to the natural angular frequency ω0. For large γ, the resonance is broad and the peak is shifted to lower frequencies. The value of γ may be so large that there is no resonance. At the resonance frequency the external force and the velocity of the particle are in phase. As a result, the power transfer (P = F·v) to the oscillator has its maximum value. At frequencies above or below the resonance value, the force and velocity are not in phase, so the power transfer is lower. 圖 Figure 15-17 The amplitude of a driven oscillator display a resonance as the angular frequency of the external agent is varied. For large damping, occurs below the natural angular frequency ω0 and resonance is broad. Even large-scale structures, such as towers, bridges, and airplanes, can oscillate. If the frequency of the driving mechanism is close to the natural frequency, the object can literally be shaken to pieces. A dramatic example of resonance is the collapse of the Tacoma Narrows Bridge in Washington. The wind flowing past the structure set the bridge into one of its natural modes of vibration, as in Fig. 15-18a. After a couple of hours the amplitude became so large that the center span disintegrated (Fig. 15-18b). In later chapters we will discuss resonance in electrical circuits, which is vital to transmission and reception of radio and TV signals. Resonance also plays a role in atomic and nuclear processes. 圖 Figure 15-18 (a) In July 1940, high winds set the Tacoma Narrows Bridge into oscillation. (b) After a couple of hours, the center span collapsed. 1.56 盪鞦韆 角動量/位能與動能 當你盪鞦韆的時候,首先要先用力衝,使它盪高起來,接著只要稍微用力就能輕 易盪著鞦韆。這種衝勁是怎麼產生效果的?若你想從靜止開始,該怎麼用力?站 著或坐在鞦韆上,用力的方式相同嗎?若鞦韆的轉動承軸非常光滑,你有可能盪 一整圈嗎?或者在高度上有某種限制?考慮一下鞦韆的兩根把手,在不同材質的 情況下,如剛硬的鐵條、鎖鍊或繩索,從靜止到某個最大高度,你必須做多少功? 1.57 士兵列隊齊步通過便橋 振盪/共振 1831 年,在英國曼徹斯特附近,有一隊騎兵以整齊的步伐通過一座吊橋,他們 的行進使橋發生擺動,最後橋垮了下來。從那時候起,部隊過橋時不再齊步通過。 這件事是怎麼發生的?真的有這種危險嗎?可能的話,做個簡單計算。 2-84 塔科馬海峽吊橋的崩塌 驅動共振/諧振盪 你可能知道美國塔科馬海峽(Tacoma Narrows)吊橋的坍塌,很多學校的物理 系都保存著整個過程的紀錄片,清楚顯現出橋的振盪以及最後的坍他情形。 橋在建造的時候就振盪的很厲害,事實上結構物的擺動甚至讓工人覺得暈眩。甚 至在橋正式通車之後,很多遠方的人特別開車過來,就為了體會吊橋振盪的刺 激。有些日子,橋的振盪可達到 5 英尺之多,橋上車子裡的乘客甚至看不到別的 車子。 然而橋最後會垮掉,至今仍然令人吃驚。在坍塌的那天早上,橋的擺動忽然停止 了,而在短暫的停頓之後,橋發生猛烈的扭力振盪(torsional oscillation) 。當時 在橋上的兩個人只好四枝著地,爬了出來。後來發現一隻狗陷在橋上,有個教授 只好在扭力振盪的節線(nodal line)上,倒退著前進(他的倒退救狗行為也被 記錄下來)。 在扭力振盪了三十分鐘之後,一塊橋面上的鐵板掉了下來。又過了三十分鐘,600 英尺長的部份橋板也掉了下來。之後,儘管扭動曾短暫停止,但旋即扭動起來, 不到幾分鐘的時間,主要的橋板就全掉下來了。 這件事並不能怪罪橋的設計者(悲劇發生後,他的事業生涯結束,不久候也去世 了),因為當時對吊橋的空氣動力行為還不十分瞭解。這件事對橋樑建築的影響 是巨大而深遠的。 在物理學上,把這座橋的坍塌當成是強迫共振(driven resonance)的實例。雖 然在那天,風並不是特別的強,但橋的振盪卻大到發生災難的程度。但是風怎麼 會弄成這樣,又如何造成這種後果?為什麼很穩定的風會造成擺動,而這擺動隨 後引發扭力振盪?既然強迫共振意味著驅動力和被驅動的物體之間某種頻率的 契合,你必須說明風如何產生這種契合頻率。 資料來源:物理馬戲團(I)天下文化出版