Transcript 幻灯片 1 - Yangtze University

Mechanics Exercise Class Ⅳ
1 Simple Harmonic Motion
x  xm cos( t   )
d 2x
2


x0
2
dt
F   kx
Velocity
Acceleration
v  xm sin( t   )
a  2 xm cos( t   )
1
E  U  K  kxm2
2
2 Energy
3 Pendulums
T  2 L
g
Simple pendulum T  2 I mgh Physical pendulum
4 Damped Harmonic Motion
x( t )  xme
bt 2 m
cos( t   )
 
k
b2

m 4m 2
5 Forced Oscillations and Resonance
d  
The greatestvm
6 Transverse and Longitudinal Waves
Sinusoidal Waves
k  2

y( x,t )  ym sin( kx  t )
v      f
k
T
7 Wave Speed on Stretched String
The average power
Pavg
v


1
 v2 ym2
2
8 Interference of Waves
y( x,t )  [ 2 ym cos
1
1
 ] sin( kx  t   )
2
2
9 Standing Waves
y( x,t )  [ 2 ym sin kx ] cos x
resonance
v
v
f  n

2L
10 Sound Waves
v B
11 Interference

(m=0.1.2…)
2m
 
L 

(2m  1)
2
Constructive interference
Destructive interference
12 The Doppler Effect
v  vD
f f
v  vS
vD
The speed of the detector
vS
The speed of the source
13 Flow of Ideal Fluids
Rv  Av =a constant
Rm  Rv =a constant
14 Bernoulli’s Equation
1 2
p  v  gy =a constant
2
1 Suppose that the two springs in figure have different spring
constants k1 and k2. Show that the frequency f of oscillation of the block
is then given by f  f 2  f 2 where f1 and f2 are the frequencies at
1
2
which the block would oscillate if connected only to spring 1 or only to
spring 2.
m
Solution Key idea:
The block –springs system forms
a linear simple harmonic oscillator,
with the block undergoing SHM.
k2
k1
o
Assuming there is a small displacement x , then the spring 1 is
Stretched for x and the spring 2 is compressed for x at the same
time. From the Hook’s law we can write
Fx  k1 x  k2 x=  ( k1  k2 )x
The coefficient of a simple spring
x
Using the Newton’s Second Low , we can obtain
d2x
m 2  ( k1  k 2 ) x
dt
d2x
m 2  (k1  k 2 ) x=0
dt
k1  k2
 
m
Thus the angular frequency is
And the frequency f of oscillation of the block is

1
f 

2 2
1

2
k1  k2
m
12   22 
f12  f12
2 A spring with spring constant k is attached to a mass m that is
confined to move along a frictionless rail oriented perpendicular to
the axis of the spring as indicated in the figure. The spring is initially
unstretched and of length l0 when the mass is at the position x = 0 m
in the indicated coordinate system. Show that when the mass is
released from the point x along the rail, the oscillations occur but
their oscillations are not simple harmonic oscillations.
Solution:
The mass is pulled out a distance x along
the rail, the new total length of the spring is
l  l02  x 2
So the x-component of the force that the spring
exerted on the mass is
1
2 2
Fx  F cos   k[(l02  x )  l0 ]cos 
From the diagram , cos 
x
x l
2
2
0
so the x-component of the force is
x
( )2  1
l0
l0 x
1
Fx  k ( x 
)  kx (1 
)  kx (1 
)
2
2
x 2
x 2
x  l0
(
) 1
( ) 1
l0
l0
2
x
For x  l0     1 the x-component of the force is
 l0 
x
1 x
Fx  kx[1  ( ) 2  1]   kx{1  [1  ( ) 2  }
l0
2 l0
k 3
 2 x
2l0
Thus when the mass is released from the
point x along the rail, the oscillations occur
but their oscillations are not simple
harmonic oscillations.
3 The figure gives the position of a 20 g block oscillating in SHM on
the end of a spring. What are (a) the maximum kinetic energy of the
block and (b) the number of times per second that maximum is
reached?
Solution:The key idea is that it’s
a SHM.
(a) From the figure, we can get
the period T and amplitude of
the system, they are
T  40ms, xm  7cm
so the angular frequency is  
2
 50 rad / s
T
The total mechanical energy of the SHM is
1
1
2 2
E  m xm   0.02kg  ( 50 rad / s )2  ( 0.07m )2  1.21J
2
2
When the block is at its equilibrium point, it has a maximum
kinetic energy, and it equals the total mechanical energy,
that is
Em  E  1.21J
(b) Because the frequency is

f 
 25Hz
2
the number of times per second that maximum is 50
4 Two sinusoidal waves , identical except for phase, travel in
the same direction along a string and interfere to produce a
resultant wave given by
y( x, t )  (3.0mm) sin(20 x  4.0t  0.82tad )
with x in meters and t in seconds. What are (a) the wavelength of
the two waves, (b) the phase difference between them , and (c)
their amplitude ?
2
(a)From
the
equation
, we can get the wavelength
k

solution

of the two waves:
2
2


 0.314m
k
20
1
1

y
(
x,t
)

[
2
y
cos

]
sin(
kx


t

)
(b) From the equation
m
2
2
we can obtain the phase difference
1
  0.82
2
(c) Because we know
ym 
2 ym cos
  1.64rad  940
1
  3.0mm
2
, their amplitude is
3.0
1.5

 2.05mm
0
1
2 cos  cos 47
2
5 A bat is flitting about in a cave , navigating via ultrasonic bleeps.
Assume that the sound emission frequency of the bat is 39000Hz.
During one fast swoop directly toward a flat wall surface , the bat is
moving at 0.025 times the speed of sound in air. What frequency does
the bat hear reflected off the wall?
Solution:
There are two Doppler shifts in this situation.
First, the emitted wave strikes the wall, so the
sound wave of frequency is
1
f f
vs
1
v
(source moving
toward the
stationary wall)
1
 39000
1  0.025
 40000 Hz
Second , the wall reflects the wave of frequency f  and reflects
it , so the frequency detected f  , will be given :
vo
f   f ( 1 
)
v
 40000( 1  0.025 )
 41000 Hz
(observer moving toward the
stationary source)
6 A device to study the suction of flowing fluid is shown in the figure.
When the fluid passing through the horizontal pipe the liquid in tank
A can be drawn upward by the stream in the pipe. Assume that the
upper tank is big enough so that the surface of the water in it does
not descend obviously when the stream flows continuously. The
height difference between the water surface in the upper tank and the
pipe is h. The cross-sectional area of the pipe at the narrow place and
the open end are S1 and S2, respectively. The density of the ideal fluid
is ρ . Show that
 S 22 
p1  p0  gh1  2   0
 S1 
p0
h C
p1 S1
A
S2
B
p0
Solution:
Choose the height of the pipe as the
reference
level
for
measuring
elevations and point O at the surface
of the water in the upper tank, point 1
at the narrow place and point 2 at the
open end of the pipe.
Applying Bernoulli’s equation to a streamline passing
through point O and 1:
1
1
p0  v02  gh0  p1  v12  gh1
2
2
As po  p1 , v0  0, ho  h, h1  0
Then
1 2
p0  p1  gh  v1
2
(1)
Applying Bernoulli’s equation to a streamline passing
through point O and 2, we have:
1 2
gh  v2
2
(2)
So v2  2 gh
Apply the equation of continuity for point 1 and 2,we get
v1S1  v2 S 2
(3)
Thus
2
 S2 
S 22
v   v2   2 2 gh
S1
 S1 
2
1
Substituting the above equation into Eq.(1) and noticing
that S1<S2, then
 S 22 
p1  p0  gh1  2   0
 S1 