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2 Energy
1 Simple Harmonic Motion
21
2 mE U K kx
22
20
d xx
dtF kx mx x cos( t )
Velocity
mv x sin( t ) 2 ma x cos( t )
Acceleration
2 IT mgh
3 Pendulums
2 LT g Simple pendulumSimple pendulum Physical pendulumPhysical pendulum
4 Damped Harmonic Motion
2bt mmx( t ) x e cos( t )
2
24
k b
m m
5 Forced Oscillations and Resonance
6 Transverse and Longitudinal Waves
my( x,t ) y sin( kx t ) Sinusoidal Waves
v fk T 2k
d mvThe greatestThe greatest
2 21
2avg mP v y
9 Standing Waves
7 Wave Speed on Stretched String v
8 Interference of Waves
1 12
2 2 my ( x,t ) [ y cos ] sin( kx t )
2 my ( x,t ) [ y sin kx ] cos x
The average powerThe average power
2
v v
f nL
resonance
12 The Doppler Effect
10 Sound Waves
11 Interference
D
S
v vf f
v v
22
(2 1)
mL
m
22
(2 1)
mL
m
Constructive interference
Destructive interference
(m=0.1.2…)(m=0.1.2…)
Bv
Sv
Dv The speed of the detector
The speed of the source
14 Bernoulli’s Equation
13 Flow of Ideal Fluids
vR Av =a constant=a constant
21
2 p v gy =a constant=a constant
m vR R =a constant=a constant
Solution Key idea:
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 where f1 and f2 are the frequencies at
which the block would oscillate if connected only to spring 1 or only to
spring 2.
2 21 2f f f
m1k 2k
The block –springs system forms The block –springs system forms a linear simple harmonic oscillator,a linear simple harmonic oscillator,with the block undergoing SHM.with the block undergoing SHM. oo xx
Assuming there is a small displacement Assuming there is a small displacement xx , then the spring 1 is , then the spring 1 is Stretched for Stretched for xx and the spring 2 is compressed for and the spring 2 is compressed for xx at the same at the same time. From the Hook’s law we can write time. From the Hook’s law we can write
x 1 2 1 2F k x k x= ( k k )x
The coefficient of a simple spring
( )
( )
2
1 22
2
1 22
d xm k k x
dt
d xm k k x=0
dt
Using the Newton’s Second Low , we can obtain Using the Newton’s Second Low , we can obtain
1 2k k
m
Thus the angular frequency is Thus the angular frequency is
And the frequency f of oscillation of the block is
1 2
2 2 2 21 2 1 1
1
2 21
2
k kf
m
f f
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 initiallyunstretched and of length l0 when the mass is at the position x = 0 m
in the indicated coordinate system. Show that when the mass isreleased 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
2 20l l x
12 2 20 0cos [( ) ]cosxF F k l x l
So the x-component of the force that the spring exerted on the mass is
2 2
0 0
320
1[1 ( ) 1] {1 [1 ( ) }
2
2
x
x xF kx kx
l l
kx
l
20
2cos
lx
x
From the diagram , so the x-component of the force is
2
00
2 2 220
00
( ) 11
( ) (1 ) (1 )( ) 1( ) 1
x
xll x
F k x kx kxxxx lll
Thus when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations.
For the x-component of the force is 12
00
l
xlx
3 3 The figure gives the position of a 20 g block oscillating in SHM on 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 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 block and (b) the number of times per second that maximum is reached?reached?
Solution:The key idea is that it’s a SHM.
(a) From the figure, we can get (a) From the figure, we can get the period the period TT and amplitude of and amplitude of the system, they are the system, they are
40 7mT ms, x cm
so the angular frequency is 2
50 rad / sT
The total mechanical energy of the SHM is
2 2 2 21 10 02 50 0 07 1 21
2 2mE m x . kg ( rad / s ) ( . m ) . J
(b) Because the frequency is
252
f Hz
the number of times per second that maximum is 50
When the block is at its equilibrium point, it has a maximum When the block is at its equilibrium point, it has a maximum kinetic energy, and it equals the total mechanical energy, kinetic energy, and it equals the total mechanical energy, that is that is
1 21mE E . J
4 Two sinusoidal waves , identical except for phase, travel in Two sinusoidal waves , identical except for phase, travel in the same direction along a string and interfere to produce a the same direction along a string and interfere to produce a resultant wave given by resultant wave given by
with with xx in meters and in meters and tt in seconds. What are (a) the wavelength of in seconds. What are (a) the wavelength of the two waves, (b) the phase difference between them , and (c) the two waves, (b) the phase difference between them , and (c) their amplitude ? their amplitude ?
( , ) (3.0 )sin(20 4.0 0.82 )y x t mm x t tad
2 20.314
20m
k
solutionsolution2
k
(a)From the equation , , we can get the wavelengthof the two waves:
0
3.0 1.52.05
1 cos 472cos2
my mm
(b) From the equation 1 1
22 2
my ( x,t ) [ y cos ] sin( kx t )
we can obtain the phase difference
10.82
2 01.64 94rad
12 cos 3.0
2my mm (c) Because we know , their amplitude is
55 A bat is flitting about in a cave , navigating via ultrasonic bleeps. A bat is flitting about in a cave , navigating via ultrasonic bleeps.
Assume that the sound emission frequency of the bat is 39000Hz.Assume that the sound emission frequency of the bat is 39000Hz.
During one fast swoop directly toward a flat wall surface , the bat is During one fast swoop directly toward a flat wall surface , the bat is
moving at 0.025 times the speed of sound in air. 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.There are two Doppler shifts in this situation.First, the emitted wave strikes the wall, so the First, the emitted wave strikes the wall, so the sound wave of frequency is sound wave of frequency is
1
1
139000
1 0 02540000
s
f fv
v
.Hz
((source movingsource moving toward thetoward the
stationary wall)stationary wall)
1
40000 1 0 025
41000
ovf f ( )
v( . )
Hz
f Second , the wall reflects the wave of frequency and reflectsit , so the frequency detected , will be given :f
((observer moving toward theobserver moving toward the stationary source)stationary source)
66 A device to study the suction of flowing fluid is shown in the figure. 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 tankWhen the fluid passing through the horizontal pipe the liquid in tankA can be drawn upward by the stream in the pipe. Assume that the 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 upper tank is big enough so that the surface of the water in it does not descend obviously when the stream flows continuously. The not descend obviously when the stream flows continuously. The height difference between the water surface in the upper tank and height difference between the water surface in the upper tank and
the the pipe is pipe is hh. The cross-sectional area of the pipe at the narrow place and . The cross-sectional area of the pipe at the narrow place and the open end are the open end are SS11 and and SS22, respectively. The density of the ideal fluid , respectively. The density of the ideal fluid
is is ρρ . Show that . Show that 012
1
22
01
S
Sghpp
A B
C1Sh
2S0p1p
0p Solution:Choose the height of the pipe as the Choose the height of the pipe as the reference level for measuring reference level for measuring elevations and point elevations and point OO at the surface at the surface of the water in the upper tank, point 1 of the water in the upper tank, point 1 at the narrow place and point 2 at the at the narrow place and point 2 at the open end of the pipe.open end of the pipe.
Applying Bernoulli’s equation to a streamline passing Applying Bernoulli’s equation to a streamline passing through point through point OO and 1: and 1:
12110
200 2
1
2
1ghvpghvp
Then (1)Then (1)
AsAs 0,,0, 101 hhhvpp oo
2110 2
1vghpp
Applying Bernoulli’s equation to a streamline passing Applying Bernoulli’s equation to a streamline passing through point through point OO and 2, we have: and 2, we have:
222
1vgh
SoSo ghv 22
Apply the equation of continuity for point 1 and 2,we getApply the equation of continuity for point 1 and 2,we get
2211 SvSv
(2)(2)
(3)(3)