國立中山大學 98 學年度普通物理(一)第三次段考
99.01.14
答題須知：
作答時，請由答題紙第六列開始填寫，並標明題號。考試時間為 19：00~21：30，20：00 以後方可交卷。請記得寫上單
位，建議以 M.K.S 制表示。
1. In Fig. 1, A uniform rope of mass m and length L hangs from a
ceiling.
(a) (5%)Show that the speed of a transverse wave on the rope
is a function of y, the distance from the lowest end, and is
m
L
y
given by v  gy .
y=0
(b) (5%)Show that the time a transverse wave takes to travel
Fig. 1
the whole length of the rope is given by t  2 L / g .
(a) The wave speed at any point on the rope is given by v =   , where  is the
tension at that point and  is the linear mass density. Because the rope is
hanging the tension varies from point to point. Consider a point on the rope a
distance y from the bottom end. The forces acting on it are the weight of the
rope below it, pulling down, and the tension, pulling up. Since the rope is in
equilibrium, these forces balance. The weight of the rope below is given by
gy, so the tension is  = gy. The wave speed is
v   gy /   gy.
【寫出 v=   的表示式給 1 分，代換出最後結果再給 4 分】
(b) The time dt for the wave to move past a length dy, a distance y from the
bottom end, is dt  dy v  dy
gy and the total time for the wave to move
the entire length of the rope is
L
t
L
0
【寫出 v 
dy
y
2
g
gy
2
0
L
.
g
dy
or dt = dy / v or dt  dy / gy 給 1 分，算出積分結果再給 4 分】
dt
2. A nylon guitar string has a linear density of 7.20 g/m and
is under a tension of 150 N. The fixed supports are
distance D = 90.0 cm apart. The string is oscillating in
the standing wave pattern shown in Fig. 2. Calculate
Fig. 2
(a) (4%)The speed of the wave
(b) (3%)The wavelength
(c) (3%)The frequency of the traveling waves whose superposition gives this
standing wave
(a) The speed of the wave is
v

150N


7.20  103 kg / m
【列至此式給 2 分】
v = 144.34 m/s  1.44×102 m/s
【算出最後結果再給 2 分，無單位扣 1 分】
(b) From the figure, we find the wavelength of the standing wave to be
 = (2/3)(90.0 cm) = 60.0 cm.
【全對(含單位)才給分】
(c) The frequency is
f 
v 1.44 102 m/s

 241Hz.

0.600 m
【全對(含單位)才給分】
3. The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00
kHz is 0.600 W/m2.
(a) (5%)What is the sound level of this wave in dB(聲音強度的基準為 I0 = 10-12
W/m2)?
(b) (5%)Determine the intensity if the frequency is increased to a 2.50 kHz while
a constant displacement amplitude is maintained.
(a) 聲音強度的基準為 I o  1012 W/m 2
本題擴音器的聲音強度為 I  0.600 W/m 2 , 其強度級β為
I
0.6
 10  log 12  10  log(6  1011 )
I0
10
  10  log
【列至此式給 4 分】
 10(log 6  11log10)  10(11  0.778)
 117.75 dB
【算出最後結果(含單位)再給 1 分】
(b) 聲波強度之表示式為 I 
1
2
 v( S max ) 2  2  v 2 f 2 S max
2
聲波位移振輻 S max 為定值時,
【寫出
2
I  f 2,
I'
f'
 ( )2
I
f
∴
I'
f'
 ( )2 給 4 分】
I
f
2
 f '
 2.5  103 
I '    I  
 0.600 W / m 2 
3
 10

 f 
= 2.52(0.600 W/m2) = 3.75 W/m2.
【註：聲波強度之符號可用 Smax , Ymax , 或 Amax 表示.】
【算出最後答案(含單位)再給 1 分】
4. A point source emits 30.0 W of sound isotropically(等向性地). A small
microphone intercepts the sound in an area of 0.750 cm2, 200 m from the source.
Calculate
(a) (3%)the sound intensity there
(b) (2%)the power intercepted by the microphone
波面
面積 A
(a)
r = 200 m
點波源的功率 P  30.0 W
點波源在均勻介質中產生球面波
P
I
半徑為 200 m 的球面波波面面積為
A = 4r2 = 4(3.14)(200)2 = 5.024×105 m2
球面波上的強度為
I
P
30.0 W

 5.97  105 W / m 2 .
5
2
A 5.024  10 m
【全對(含單位)才給分】
波面
面積 A
r = 200 m
(b)
耳機的面積為 A = 0.75×10-4 m2
耳機接收到的聲波強度為
I
耳機
面積 A
P = I A = (5.97×10-5 W/m2)(0.75×10-4 m2) = 4.48×10-9 W or 4.48 nW
【全對(含單位)才給分】
耳機
P
5. (5%)The maximum pressure amplitude Pm that the human ear can tolerate in
loud sounds is about 28 Pa (1 Pa = 1 N/m2). What is the displacement amplitude
Sm for such a sound in air of density  = 1.21 kg/m2, at a frequency of 1000 Hz
and a speed of 343 m/s?
聲波可用位移波與壓力波表示
位移波之波函數若為 S ( x, t )  Sm cos(kx  t )
則壓力波之波函數為
P( x, t )  Pm cos(kx  t   2)  Pm sin(kx  t )
上式中 P ( x, t ) 也可用 P( x, t ) 表示, Pm 也可用 Pm 表示
壓力波振幅 Pm 與位移波振幅的關係為
Pm  k  v 2 S m   v S m 【寫出此式即給 4 分】
Sm 
Pm
P
Pm
 m 
2
 v v  v (2 f )
28 N/m2
(343 m/s)(1.21 kg/m3 )(2 )(1000 Hz)
 1.1105 m  11  m
【算出最後結果(含單位)再給 1 分】

vA = 0
6. (a) (5%)In Fig. 3, Two whistles A and B, each
vo = 30 m/s
vB = 60 m/s
have frequency of 680 Hz, A is stationary, and
A
O
B
B is moving toward right at velocity 60 m/sec,
Fig. 3
an observer is between A and B, moving
toward the right with velocity v0 = 30 m/sec. Take the velocity sound in air as 340
m/sec. What is the beat frequency heard by the observer?
(b) (5%)A bat (蝙蝠) emits sound at frequency f = 38000 Hz, and fly directly
toward a flat wall of a cave with velocity 1/20 times the speed of sound in air,
what frequency does the bat hear reflected off the wall.
vA = 0
A
(a) O 聽到 A 之頻率：
vo = 30 m/s
O
vB = 60 m/s
B
 v  v0 
340  30 
310
  680
f A  f 0 
 620 Hz
  680 
v

v
340

0
340


s 

O 聽到 B 之頻率：
 v  v0 
 340  30 
370
 340  30 
  680
f B  f 0 
 629 Hz 【or f B  680
】
  680 
400
 340  60 
  340  60 
 v  vs 
∴fb (拍頻率) = fB－fA = 629－620 = 9 Hz
【算出 fA 給 2 分，算出 fB 給 2 分，算出 fb 再給 1 分，最後答案無單位扣 1 分】
(b) 在 wall 接到 bat 之頻率
f1  f 0
vair = v
v  v0
v0
v
 f0
 f0
【2 分】
v  vs
v  v / 20
v  v / 20
bat
vb = v/20
bat 接到 wall 反射回來之頻率
f 2  f1
 v  v / 20
v  v / 20
 f1
【2 分】
v0
v
v
v  v / 20
21
21

  v  v / 20 
 f0
 38000 
 42000 Hz

  f0
v
v  v / 20
19
19
 v  v / 20  

∴ f2  f0 
【算出 f1 給 2 分，算出 f2 給 2 分，算出最後答案(含單位)再給 1 分】
【利用投射方式直接列出 f 2  f 0
v  v / 20
也可以】
v  v / 20
f1
7. (a) (5%)A pipe with both end open, has length L. What is the resonant frequency
of the third harmonic? (Assume the sound speed in air is v.)
(b) (5%)In the case (a), if the pipe only one end open, find the resonant frequency
of the first harmonic.
(a) both end open (two open end)
f 
v


v
nv

2L / n 2L
for n = 3
f 
3v
.
2L
【答案對即給全分】
(b) one end open
for n = 1 (first harmonic)
L
1
v
v
 ，∴ f  
4
 4L
【答案對即給全分】
8. (10%)A sample of an ideal gas goes through
the process shown in Fig. 4. From A to B, the
process is adiabatic; from B to C, it is isobaric
with 100 kJ of energy entering the system by
heat. From C to D, the process is isothermal;
from D to A, it is isobaric with 150 kJ of
energy leaving the system by heat. Determine
the difference in internal energy Eint,B – Eint,A.
＊ HALLIDAY 用書之定義 ＊
Fig. 4
＊ SERWAY 用書之定義 ＊
BC 過程
BC 過程
WBC = PB(VC－VB)=3 atm (0.400－0.0900) m3 = 94.2 kJ
WBC = －PB(VC－VB)=－3 atm (0.400－0.0900) m3 = －94.2 kJ
【算出 BC 過程的功給 1 分】
【算出 BC 過程的功給 1 分】
Eint = Q－W
Eint = Q＋W
Eint,C－Eint,B = (100－94.2) kJ = 5.79 kJ
Eint,C－Eint,B = [100＋(-94.2) kJ] = 5.79 kJ
(內能也可以 U 表示)
(內能也可以 U 表示)
【算出 BC 過程的內能變化給 2 分】
【算出 BC 過程的內能變化給 2 分】
CD 過程
CD 過程
Since T is constant,
Since T is constant,
Eint,D  Eint,C  0
Eint,D  Eint,C  0
【算出 CD 過程的內能變化給 2 分】
【算出 CD 過程的內能變化給 2 分】
DA 過程
DA 過程
WDA = PD(VA－VD) = 1 atm (0.200－1.20) m = －101 kJ
WDA = －PD(VA－VD)=－1.00 atm (0.200－1.20) m3 = 101 kJ
【算出 DA 過程的功給 1 分】
【算出 DA 過程的功給 1 分】
Eint = Q－W
Eint = Q＋W
Eint,A－Eint,D = [-150－(-101)] kJ = －48.7 kJ
Eint,A－Eint,D = (-150 + 101) kJ = －48.7 kJ
3
(內能也可以 U 表示)
(內能也可以 U 表示)
【算出 DA 過程的內能變化給 2 分】
【算出 DA 過程的內能變化給 2 分】
Now,
Now,
Eint,B－Eint,A = －[(Eint,C－Eint,B)＋(Eint,D－Eint,C)＋(Eint,A－Eint,D)]
Eint,B－Eint,A = －[(Eint,C－Eint,B)＋(Eint,D－Eint,C)＋(Eint,A－Eint,D)]
Eint,B－Eint,A = －[5.79 kJ＋0－48.7 kJ] =
Eint,B－Eint,A = －[5.79 kJ＋0－48.7 kJ] =
42.9 kJ
【算出 AB 過程的內能變化(含單位)給 2 分】
42.9 kJ
【算出 AB 過程的內能變化(含單位)給 2 分】
【直接利用 W   PdV 算出 AB 過程的功，再代入熱力學第一定律解內能變化也可以】
9. (10%)A solar energy collector (solar cooker) consists of a curved reflecting
surface that concentrates sunlight onto the object to be warmed. The solar power
per unit area reaching the Earth’s surface at the location is 600 W/m2. The cooker
faces the Sun and has a diameter of 0.600 m. Assume that 40.0% of the incident
energy is transferred to 0.500 kg of water in an open container, initially at 20.0C.
How long does it take to completely boil away the water?
(Ignore the heat capacity of the container.)
(The specific heat of water: 4186 J/kg℃)
(The latent heat of vaporization of water: 2.26×106 J/kg)
The power incident on the solar collector is


2
Pi  IA  600 W m 2   0.300 m    170 W


【算出到達 collector 的功率給 2 分】
. The total energy required to increase the temperature of the water to the boiling
point and to evaporate it is Q  cm T  m LV :
【列出 20℃的水完全沸騰所需熱量之式子給 2 分】
Q  0.500 kg  4186 J kg C   80.0C   2.26  106 J kg  1.30  106 J
【算出熱量值給 2 分】
For a 40.0% reflector, the collected power is Pc  67.9 W
The time interval required is t
Q 1.30  106 J

 5.31 h
Pc
67.9 W
【列出t = Q/Pc 給 2 分，算出最後答案(含單位)再給 2 分】
10. (4%)An ideal gas expands at a constant temperature T from an initial volume Vi to
a final volume Vf. Derive how much work done during the expansion?
＊ HALLIDAY 用書之定義 ＊
＊ SERWAY 用書之定義 ＊
(i )W  V pdV
(i )W   V pdV
nRT
代入
V
V nRT
V
(iii)W  V
dV  nRT [ln V ]V
V
V
( IV )W  nRT ln f
Vi
(ii)將p 
Vf
Vf
i
i
(ii)將p 
f
i
【每一個步驟 1 分】
f
i
nRT
代入
V
V nRT
V
(iii)W   V
dV  nRT [ln V ]V
V
V
( IV )W  nRT ln f
Vi
f
i
f
i
【每一個步驟 1 分】
11. (3%)The molar mass of nitrogen molecules (N2) is 0.028 kg/mol. Compute the rms
speed of a nitrogen molecule at 20.0°C. (The ideal gas constant R：8.31 J/molK)
vrms 
3RT
3(8.31J / mol  K )(273  20) K

 510.75m / s
M
0.028kg / mol
【寫出公式即給 1 分，算出最後結果(含單位)再給 2 分】
12. The temperature of 2.00 mol of an ideal monatomic gas is raised from 300.0 K to
315.0 K at constant volume.
(a) (3%)What is the average kinetic energy per atom of the initial state at 300 K?
(b) (3%)What is the work W?
(c) (3%)What is the energy transferred as heat Q?
(d) (2%)What is the change Eint in the internal energy of the gas?
(e) (2%)If the temperature of 2.00 mol of an ideal monatomic gas is raised 15.0
K at constant pressure, what is the work W during the process?
(The ideal gas constant R：8.31 J/molK；Avogadro’s number NA：6.02×1023 mol-1)
3
3
kT  (1.38  1023 J / K )(300K )  6.21 1021 J
2
2
(a) Eavg 
(b)W  V pdV  p(V f  Vi )  0
Vf
i
3
3
(c)Q  nCv T (or nRT )  (2mol )( )(8.31J / mol  K )(15K )  373.95J
2
2
3
3
(d )E int  nRT (or nCv T )  (2mol ) (8.31J / mol  K )(15K )  373.95J
2
2
(e) W  
Vf
pdV  pV  nRT  (2mol )(8.31J / mol  K )(15K )  249.3J
Vi
or
W   V pdV   pV  nRT  (2mol )(8.31J / mol  K )(15K )  249.3J
Vf
i
【寫出公式即給 1 分，算出最後結果(含單位)再給全分】