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```Mechanics Exercise Class в…Ј
1 Simple Harmonic Motion
2
x пЂЅ x m cos( пЃ· t пЂ« пЃ¦ )
d x
F пЂЅ пЂ­ kx
dt
Velocity
2
Acceleration
v пЂЅ пЂ­ пЃ· x m sin( пЃ· t пЂ« пЃ¦ )
a пЂЅ пЂ­ пЃ· x m co s( пЃ· t пЂ« пЃ¦ )
2
E пЂЅU пЂ« K пЂЅ
2 Energy
2
пЂ«пЃ· xпЂЅ0
1
2
2
kx m
3 Pendulums
T пЂЅ 2пЃ°
L
g
Simple pendulum
T пЂЅ 2пЃ°
I
m gh
Physical pendulum
4 Damped Harmonic Motion
x( t ) пЂЅ x m e
пЂ­ bt 2 m
co s( пЃ· п‚ў t пЂ« пЃ¦ )
пЃ·п‚ў пЂЅ
k
m
5 Forced Oscillations and Resonance
пЃ·d пЂЅ пЃ·
The greatestv m
6 Transverse and Longitudinal Waves
Sinusoidal Waves
k пЂЅ 2пЃ°
пЃ¬
y( x ,t ) пЂЅ y m sin( kx пЂ­ пЃ· t )
vпЂЅпЃ·
k
пЂЅпЃ¬
T
пЂЅ пЃ¬f
пЂ­
b
2
4m
2
v пЂЅ
7 Wave Speed on Stretched String
The average power
Pavg пЂЅ
1
2
пЃґ
пЃ­
пЃ­vпЃ· ym
2
2
8 Interference of Waves
y п‚ў( x ,t ) пЂЅ [ 2 y m co s
1
пЃ¦ ] sin ( kx пЂ­ пЃ· t пЂ«
2
1
пЃ¦)
2
resonance
9 Standing Waves
y п‚ў( x ,t ) пЂЅ [ 2 y m sin kx ] cos пЃ· x
f пЂЅ
v
пЃ¬
пЂЅn
v
2L
10 Sound Waves
v пЂЅ
11 Interference
B
пЃІ
(m=0.1.2вЂ¦)
пѓ¬п‚±2mпЃ°
пЃ„пЃ¦ пЂЅ
пЃ„L пѓ­
пЃ¬
пѓ®п‚±(2m пЂ« 1)пЃ°
Constructive interference
2пЃ°
Destructive interference
12 The Doppler Effect
fп‚ўпЂЅ f
v п‚± vD
vD
The speed of the detector
v п‚± vS
vS
The speed of the source
13 Flow of Ideal Fluids
R v пЂЅ Av =a constant
R m пЂЅ пЃІ R v =a constant
14 BernoulliвЂ™s Equation
pпЂ«
1
2
пЃІ v пЂ« пЃІ 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 пЂЅ f 2 пЂ« 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.
Solution Key idea:
The block вЂ“springs system forms
a linear simple harmonic oscillator,
with the block undergoing SHM.
m
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 HookвЂ™s law we can write
Fx пЂЅ пЂ­ k 1 x пЂ­ k 2 x= пЂ­ ( k 1 пЂ« k 2 )x
The coefficient of a simple spring
x
Using the NewtonвЂ™s Second Low , we can obtain
2
m
d x
dt
2
пЂЅ пЂ­ (k1 пЂ« k 2 ) x
2
m
d x
dt
2
пЂ« ( k 1 пЂ« k 2 ) x= 0
k1 пЂ« k 2
пЃ· пЂЅ
Thus the angular frequency is
m
And the frequency f of oscillation of the block is
f пЂЅ
пЂЅ
пЃ·
2пЃ°
1
2пЃ°
пЂЅ
1
k1 пЂ« k2
2пЃ°
m
пЃ·12 пЂ« пЃ· 22 пЂЅ
f1 пЂ« f1
2
2
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 пЂЅ
l0 пЂ« x
2
2
So the x-component of the force that the spring
exerted on the mass is
1
F x пЂЅ F cos пЃ± пЂЅ пЂ­ k [( l 0 пЂ« x ) 2 пЂ­ l 0 ] cos пЃ±
2
2
cos пЃ± пЂЅ
From the diagram ,
x
x пЂ«l
2
2
0
so the x-component of the force is
(
Fx пЂЅ пЂ­k ( x пЂ­
l0 x
x пЂ«l
2
1
) пЂЅ пЂ­ kx (1 пЂ­
2
0
(
x
) пЂ«1
(
For
F x п‚» пЂ­ kx [1 пЂ­
пЂЅпЂ­
k
2
2 l0
x
2
x
)
) пЂ«1
2
l0
l0
пѓ¦ x
x пЂјпЂј l 0 пѓћ пѓ§пѓ§
пѓЁ l0
) пЂ«1
l0
) пЂЅ пЂ­ kx (1 пЂ­
2
x
2
пѓ¶
пѓ· пЂјпЂј 1
пѓ·
пѓё
(
x
l0
the x-component of the force is
) пЂ« 1] пЂЅ пЂ­ kx {1 пЂ­ [1 пЂ­
2
1
(
x
) пЂ­
2
}
2 l0
3
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 itвЂ™s
a SHM.
(a) From the figure, we can get
the period T and amplitude of
the system, they are
T пЂЅ 40 m s , x m пЂЅ 7 cm
so the angular frequency is пЃ·
пЂЅ
2пЃ°
пЂЅ 50 пЃ° rad / s
T
The total mechanical energy of the SHM is
E пЂЅ
1
2
mпЃ· x пЂЅ
2
2
m
1
2
п‚ґ 0 . 02 kg п‚ґ ( 50 пЃ° rad / s ) п‚ґ ( 0 . 07 m ) пЂЅ 1 . 21 J
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
E m пЂЅ E пЂЅ 1 . 21 J
(b) Because the frequency is
f пЂЅ
пЃ·
2пЃ°
пЂЅ 25 H z
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 п‚ў ( x , t ) пЂЅ (3 .0 m m ) sin ( 2 0 x пЂ­ 4 .0 t пЂ« 0 .8 2 ta d )
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 ?
2пЃ°
(a)From
the
equation
, we can get the wavelength
k
пЂЅ
solution
пЃ¬
of the two waves:
пЃ¬ пЂЅ
2пЃ°
пЂЅ
k
2пЃ°
20
пЂЅ 0 .3 1 4 m
y п‚ў( x ,t ) пЂЅ [ 2 y m co s
(b) From the equation
we can obtain the phase difference
1
2
ym пЂЅ
2 y m co s
3.0
2 cos
1
2
пЃ¦ ] sin ( kx пЂ­ пЃ· t пЂ«
2
1
2
пЃ¦ пЂЅ 0 .8 2
(c) Because we know
1
пЂЅ
пЃ¦
1
пЃ¦ пЂЅ 3 .0 m m
2
1.5
cos 47
0
, their amplitude is
пЂЅ 2.05mm
пЃ¦)
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
f п‚ў пЂЅ
(source moving
1
f
1пЂ­
toward the
stationary wall)
vs
v
пЂЅ 39000
1
1 пЂ­ 0 .0 2 5
пЂЅ 40000 H z
Second , the wall reflects the wave of frequency f п‚ў and reflects
it , so the frequency detected f п‚ўп‚ў , will be given :
f п‚ўп‚ў пЂЅ f п‚ў( 1 пЂ«
vo
)
v
пЂЅ 4 0 0 0 0( 1 пЂ« 0 .0 2 5 )
пЂЅ 41000 H z
(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 ПЃ . Show that
2
пѓ¦
S2 пѓ¶
p 1 пЂ­ p 0 пЂЅ пЃІ gh пѓ§пѓ§ 1 пЂ­ 2 пѓ·пѓ· пЂј 0
S1 пѓё
пѓЁ
p0
h
C
p 1 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 BernoulliвЂ™s equation to a streamline passing
through point O and 1:
p0 пЂ«
1
2
пЃІ v 0 пЂ« пЃІ gh 0 пЂЅ p1 пЂ«
2
As p o пЂЅ p1 , v 0 пЂЅ 0 , ho пЂЅ h , h1 пЂЅ 0
Then
p 0 пЂ­ p1 пЂЅ пЃІ gh пЂ­
1
2
пЃІ v1
2
1
2
пЃІ v1 пЂ« пЃІ gh 1
2
(1)
Applying BernoulliвЂ™s equation to a streamline passing
through point O and 2, we have:
пЃІ gh пЂЅ
1
2
So v 2 пЂЅ
пЃІv2
2
(2)
2 gh
Apply the equation of continuity for point 1 and 2,we get
v1 S 1 пЂЅ v 2 S 2
(3)
Thus
2
2
пѓ¦ S2
пѓ¶
S2
v пЂЅ пѓ§пѓ§
v 2 пѓ·пѓ· пЂЅ 2 2 gh
S1
пѓЁ S1
пѓё
2
1
Substituting the above equation into Eq.(1) and noticing
that S1<S2, then
2
пѓ¦
S2 пѓ¶
p 1 пЂ­ p 0 пЂЅ пЃІ gh пѓ§пѓ§ 1 пЂ­ 2 пѓ·пѓ· пЂј 0
S1 пѓё
пѓЁ
```
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