Download 九十七學年第一學期普通物理B 第二次段考試題 [Young12

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九十七學年第一學期 普通物理 B 第二次段考試題
[Young12-16]
2008/12/5, 8:20AM – 9:50AM
0.【5%】依下面說明在答案卷上作答者，可得 5 分
(i) 答案卷第一張為封面，第一張正反兩面不要作答。
(ii) 由第二張紙開始算起，第一頁依空格號碼順序寫下所有填充題答案，寫在其他頁不記分。
(iii) 計算題之演算過程與答案依題號順序寫在第二頁以後，每題從新的一頁寫起。
注意：單位沒寫一格扣一分
Part I.

填充題 (每格 3 分, 共 75 分)
Two spy satellites, C1 and C2, run above Taiwan 8 times (C1) and 27 times (C2) a day. The ratio
of the radii of two satellite orbits (r1/r2) is
orbits is (v1/v2)

(2)
(1)
. The ratio of the escape velocities at their current
.
Block A in Fig. 1 hangs by a cord from spring balance D and is submerged (淹沒)in a liquid C
contained in beaker B. The mass of the beaker is 1.00 kg; the mass of the liquid is 1.8 kg. The
balance D reads 2.5 kg, and the balance E reads 8.00 kg. The volume of block A is 4.0010-3 m3. If
block A is pulled up out of the liquid, balance D reads
(3) kg. The density of liquid is
(4)
3
kg/m .

In desert, the temperature can go up to 51 o C in day time and down to -17o C during night. Therefore, the wavelength of
sound varies due to the variation of speed of sound. The wavelength ratio day/night is
(5)
for
the same frequency.

The maximum pressure for 20dB sound intensity is 310-4 Pa. For 60dB sound intensity,
the maximum pressure is

.
The displacement of an oscillating object as a function of time is shown in Fig. 2. (a) the
frequency is
frequency

(6)
(7)
(10)
. (b) the amplitude is
(8)
(c) the period is
(9)
(d) the angular
.
Two identical (完全相同) stars with mass M orbit around their center of mass. Each orbit is circular and has radius R, so
that two stars are always on the opposite sides of the circle. (a) The period of the orbit is
be required to separate the two stars to infinity?

(12)
(11)
.
A piano tuner (調音師) stretches a steel piano wire with a tension of 900 N. The steel wire is 0.30 m long and has a mass of
3.00 g (a) What is the frequency of its fundamental mode of vibration?
(13)
. (b) What is the number of the highest
harmonic that could be heard by a person who is capable of hearing frequencies up to 10 kHz?


(14)
.
2
2
A 100 Hz-wave, which propagates in a string (=10 g/m), can be described as a wave equation:  y  0.64(sec 2 / m2 )  y
2
2
x
with an amplitude A = 1 mm. The wavelength of the wave is
(16)
. (b) How much energy would
. The average power of the wave is
(17)
(15)
m. The maximum speed of a particle of the string is
.
Figure 3 shows a Venturi meter, used to measure flow speed in a pipe. The flow speed v1 is
(18) in terms of the cross-sectional areas A1 and A2 and the difference in height h of the liquid
levels in the two vertical tubes.
t

(19)

If Newton’s gravitational law is needed to be modified as
Fg  G
m1m2
, which one of Kepler’s law will still stay valid?
r 2.001
.
Water stands at depth H in a large open tank whose side walls are vertical. A hole is made in
one of the walls at a depth h below the water surface. (a) Find R =
(20)
(b) How far above
the bottom of the tank could a second hole be cut so that the stream emerging from it could have
the same range as for the first hole

(21)
. (express your answer in terms of h and H)
For a forced oscillation with damping, the amplitude A vs. the driving frequency d is
shown in Fig. 5. Each curve, from a to e has different damping constant b. (a) the natural frequency of
the oscillator is
(22)
kHz (b) the most lightly damped oscillation is the curve
(23)
. (choose one
from a to e)

One taut string under a tension force F produces a note of fundamental frequency f0. Another
identical string under a slightly different tension force F+F produce a fundamental frequency f0’. The
beat frequency of f0 and f0’ is

(24)
in terms of f0, F and F. (note:
1 x  1
x
, for small x)
2
A directional loudspeaker aims a sound wave at a wall. You hear no sound at all at a distance of 2 m from the wall. What is
the lowest possible frequency of loudspeaker?
(25)
Hz. (note: speed of sound =344m/s)
Part II 計算題: 須寫下詳細的計算過程 (共 30 分)
1.
Suppose we drill a hole through the earth (radius RE, mass mE) along the diameter and drop a
mail pouch (郵袋) (mass m) down the hole with no initial velocity. r (<RE) is the distance from
pouch to the center of earth. (a) Find the potential energy of the pouch U(r) and (b) the magnitude
of the gravitational force F(r). (c) The pouch becomes a simple harmonic oscillator. Find the
angular frequency  and (d) the maximum velocity. (3 分,2 分,3 分,2 分)
2.
A U-shaped tube with a horizontal portion of length l contains a liquid. (a) What is the
difference in height between the liquid columns in the vertical arms if the tube has an acceleration
a toward the right? (b) If the tube is mounted on a horizontal turntable rotating with angular speed
 with one of the vertical arm on the axis of rotation? (5 分,5 分)
3.
A continuous triangular wave on a taut string travels in the positive x-direction with speed v. The tension in the string is F,
and the linear mass density of the string is . At t = 0, the shape of the pulse is given by
h
y(x,0)=
h
L  ( x  2nL)
L
L  ( x  2(n  1) L)
L
for 2nL<x<(2n+1)L,
for (2n+1)L<x<(2n+2)L, where n=0, 1, 2, 3…
(a) Draw the pulse at t = 0 (b) Determine the wave function y(x,t) at all times t (c) Find the average power of the wave.(3 分,3
分,4 分)
A
Part I Answer Sheet
(1)
(2)
(3)
(4)
(5)
(6)
9/4
(7)
0.1 Hz
(8)
10 mm
(9)
10 sec
(10)
0.2  sec-1
2/3
7.7
1.3103
9/8
310-2Pa
4R 3 / 2
(11)
GM
(12)
GM 2
4R
(13)
500 Hz
(14)
20
(15)
1.2510-2
(16)
0.2  m/sec
(17)
2.4710-3 W
2 gh
( A1 / A2 ) 2  1
(18)
(19)
(20)
The second one
(21)
h
(22)
1
(23)
e
(24)
(25)
2 h( H  h)
f0 (
86
F
)
2F
A
Part II Answer Sheet
1.
(a)The density of earth is

3M
4R 3
Considering a sphere with a radius of r '
For r '  r (pouch inside the sphere)
The potential energy is
dU  
GmdM
, where dM   4r ' 2 dr '
r'
R
thus U inside   
r
Gm
4r ' 2 dr '  2Gm( R 2  r 2 )
r'
For r '  r (pouch outside the sphere)
The potential energy is
dU  
GmdM
r
r
thus U outside   
0
Gm
4
4r ' 2 dr '   Gmr 2
r
3
U  U inside  U outside  2Gm( R 2  r 2 
(b) F  
2 2
GmM
r )
(3R 2  r 2 )
3
3
2R
U
GmM

r
r
R3
(c) F  ma  
GmM
r
R3
(d) vmax  A  R
a
GM
r   2 r
3
R

GM
R3
GM
GM

3
R
R
2.
(a)The net force on the horizontal part of the fluid is
a
g
ρg ( yL  yR ) A  ρAla, or, ( yL  yR )  l.
(b)The center of mass has a radial acceleration of magnitude arad   2l/2, and so the
difference in heights between the columns is
( 2l/2)(l/g )   2l 2 /2 g .
3.
(a)
(b)
x  x  vt
y(x,t)=
h
h
L  [( x  vt)  2nL]
L
L  [( x  vt)  2(n  1) L]
L
for 2nL<(x-vt)<(2n+1)L,
for (2n+1)L<(x-vt)<(2n+2)L, where n=0, 1, 2, 3…
(c) The average power of entire wave equals that of one period.
At
n=0
y(x,t)=
h
0<(x-vt)<2L
h
L  [( x  vt)]
L
 [ L  ( x  vt)]
L
for 0<(x-vt)<L,
for L<(x-vt)<2L
y y Fvh2
P  F
=
=constant
x t
L2
Fvh2
Pav  2
L