Download 彈簧水平振和垂直振，週期會不會一樣？

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
At   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
At   A0 e 0.346 t
(c) The critical damping constant is
  2m0  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 cose 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