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Unformatted text preview: Waves CHAPTER 14
WAVES and OSCILLATIONS In physics,we come across three main types of waves:
Mechanical waves: water waves, sound waves, seismic
waves, waves on a string ... • Waves
o Mathematical treatment of waves
o Doppler effect Electromagnetic waves: visible and ultroviolet light,
radio and television waves, microwaves, xrays and radar • Superposition
o Interference
o Standing waves waves. All em waves travel in vacuum with a speed of • Simple harmonic motion
o Oscillating mass on a spring
o Physical pendulum
o Simple pendulum Matter waves: the waves associated with electrons, c = 2.99792458 × 108 m s. protons and other fundamental particles, atoms and
molecules.
However, all waves have features in common. 1 A wave is caused by a disturbance, for example, when a Rather than concentrate on the properties of single string is flipped upanddown, pulses travel down the pulses, we will study periodic trains of pulses, i.e., string as a wave, as shown in (a). periodic waves. Here are examples of periodic
longitudinal and transverse waves. Notice that in both (a) cases, the particles move only backandforth or upanddown; there is no overall displacement of the points, i.e., Wave pulses can be produced by flipping a spring backandforth along its length, as shown in (b). (b) no net flow.
http://paws.kettering.edu/~drussell/Demos/waves/wavemotion.html A sound wave is a longitudinal wave; the air molecules
simply move backandforth in the direction of
propagation. The waves on a plucked string is an The motion of the disturbance is both cases is from left example of a transverse wave. to right, but in (a) the disturbance is transverse, i.e.,
perpendicular to the direction of wave propagation, and
in (b) the disturbance is longitudinal , i.e., parallel to the DISCUSSION PROBLEM [14.1]: direction of wave propagation. Notice that we can
associate a speed of propagation (left to right) with each When a breeze blows on a field of corn a wave is seen to type. pass across the field. What type of wave is it? 2 For convenience we will concentrate on transverse Mathematical description of periodic waves waves but the analysis is the same for longitudinal Even though longitudinal and transverse wave appear waves. Consider the ripples (waves) in a pond caused by rather different, they can be described mathematically in a small stone being the same way, for example as periodic, sinusoidal dropped into the functions. water ... the waves
move outward but a
cork in the water bobs
up and down.
x y
ρ y x x λ A x On the left is a snapshot of a sound wave in which the −A regions of higher density, i.e., where the atoms are closer Vertical and horizontal illustrations of the wave at
some instant in time. together, are visible. On the right is a snapshot of a
water wave in a ripple tank in which the troughs and To describe the “disturbance” of such a wave we need crests are visible. two variables, t and x. 3 y λ A x
−A y
• • • t=0
x
t = 1s
t = 2s xo Shown is a snapshot of the wave at some time t. It is
described by the general equation: y ( t , x ) = A cos(ωt − kx ) ,
where A is the amplitude. This expression describes a t=0 • • y
• t = 1s
t = 2s
y( x, t) = cos(ωt + kx) y( x, t) = cos(ωt − kx) We see that k > 0 corresponds to a wave traveling to the
right ( + x direction), and k < 0 corresponds to a wave
traveling to the left ( − x direction). traveling wave propagating along the xaxis. At some
time t, as shown here, the disturbance varies sinusoidally y t=0 x with position, x. Similarly, at some point, x o say, the
disturbance y ( t ) varies sinusoidally with time (simple
harmonic motion). The parameter k = 2π λ is called the
wavevector, where λ is the wavelength. The product kx
is called the phase angle ( = δ ). Also, ω is the angular
frequency of the wave given by ω = 2πf , and T = 1f is
the periodic time of the wave. t = 1s • • • t = 2s
y( x, t) = sin(ωt − kx) Also, we could as easily used
y ( x, t ) = A sin(ωt − kx ) as our wave equation. 4 y y λ ∆x A x x
wave at t −A wave at t + ∆t To find the speed of a wave, we take two snapshots at a Note that the disturbance at some fixed time t o, time interval ∆t apart. If the wave (i.e., the red dot) y ( t o , x ) = y ( t o , x ± nλ ), i.e., the wave is reproduced at displacements of nλ , travels a distance ∆x in that time, then the speed of the where n is an integer. wave is y v = ∆x ∆t . T A Since the disturbances (y) are equal
t −A T = 1 f = 2π ω Similarly, the disturbance at some fixed position x o,
y ( t , x o ) = y ( t ± T, x o ) = y ( t ± m 2π ω , x o ),
i.e., the wave is reproduced after time intervals of
m 2π ω , where m is an integer. y ( t , x ) = y ( t + ∆t , x + ∆ x ) ,
so ωt − kx = ω ( t + ∆t ) − k ( x + ∆x ),
i.e., ω∆t = k∆x ,
∴ v = ω k.
Note also v = ω k = 2πf 2π λ = fλ .
The velocity of a fixed point on a wave (such as the red
dot) is called the phase velocity. 5 (a) The general wave equation for a wave traveling in the
+ x direction is y ( x, t ) = A cos(ωt − kx ) or y ( x, t ) = A sin(ωt − kx ).
In this case ω = 3.50s−1 and k = −2.20 m−1. Hence k < 0
so the wave is moving to the left, i.e., in the − x direction.
Question 1: A wave on a string is described by y ( x, t ) = 0.03 sin(2.20 x + 3.50 t ) , (b) We have k = 2π λ , i.e., λ = 2π k = 2π 2.20 = 2.86 m. where the parameters are in SI units. (a) In what Also, ω = 2πf , i.e., f = ω 2π = 3.50 2π = 0.557Hz, direction is the wave moving? (b) Find the wavelength, and T = 1f = 10.557 = 1.80s. frequency and period of the wave. (c) What is the
maximum displacement, (d) what is the speed of (c) The maximum displacement is the amplitude (A), i.e., propagation of the wave, and (e) what is the maximum 0.03m. transverse speed of a point on the wave?
(d) The speed of propagation is v = ω k = 3.50 2.20 = 1.59 m/s.
(e) The transverse speed of any point on the wave is
v y = dy dt . 6 If y = A sin(ωt − kx ),
dy = Aω cos(ωt − kx ) .
dt
The maximum occurs when the cos function is ±1,
= Aω = 0.03 × 3.50 = 0.105m/s.
i.e., dy dt
max () Question 2: A wave on a string is described by
x
y ( x, t ) = 0.12 sin π 4 t + , 8
where the parameters are in SI units. What is (a) the
displacement, (b) the transverse speed, and (c) the DISCUSSION PROBLEM [14.2]:
The dimension of the wavevector k is [L]−1 so its unit is transverse accceleration, at t = 0.20s of the point located
at x = 1.60 m? (d) What is the velocity of propagation of
the wave? often given as m−1. Technically, that is not correct. So,
what is the correct unit for k? 7 x
(a) the displacement is y ( x, t ) = 0.12 sin π 4 t + . 8
∴ y (1.60 m, 0.20s) = 0.12 sin π(0.80 + 0.20) = 0. (b) the transverse speed of a point is dy dt . For the wave y ( x, t ) = A sin(ωt + kx ),
dy = Aω cos(ωt + kx ) ,
dt
x
i.e., dy dt = 0.12 × 4π cos π 4 t + 8 = 0.12 × 4π cos π(0.80 + 0.20) = −0.12 × 4π = −1.51m/s.
2 (c) the transverse acceleration is d y 2
dt x
= − Aω 2 sin(ωt + kx ) = −0.12 × (4π )2 sin π 4 t + 8 (d) the propagation speed of the wave is v =
= ω
k 4π
= 32.0 m/s.
π
8 () The velocity of a wave on a string is also related to the
tension in the string (T) and the mass of the string per
unit length ( µ ), viz:
v= T
,
µ where T is measured in Newtons and µ is in kg/m. Thus,
for two strings A and B, with µ A > µ B, for a given
tension (T), v A < v B. This is an expected result because
the inertia associated with A is greater than with B and
so it “moves” more slowly. = −0.12 × (4π )2 sin π(0.80 + 0.20) = 0 . 8 Sound waves require a medium in which to travel. As a Sound waves also transmit energy: result, the property of a particular sound wave, such as
its speed of propagation, will depend on the mass and the
spacing between atoms and the force between atoms. WBM10VD1.MOV 2so
WBM02AN1.MOV If the amplitude of the longitudinal oscillations about the
The speed of sound waves depends on temperature. We equilibrium position is so, then the energy density of a state, without proof, that the ratio of the speeds v1 and sound wave, i.e., the energy per unit volume, is
1
η = ρω 2 (so )2 ,
2 v 2 at two different absolute temperatures T1 and T2 is v2
T
= 2,
v1
T1
where T is in Kelvins so T = t o + 273.
C where ρ is the density of the medium and ω ( = 2πf ) is
the angular frequency of the wave. 9 Doppler effect
If a source of waves and a receiver are moving relative
to each other, the received frequency is different from
the frequency of the source. If the source and receiver
are moving closer together (farther apart), the received
frequency is greater (smaller) than the source frequency.
This is called the Doppler effect. If f s is the frequency of the source and v is the speed of
sound relative to still air, then the distance between
successive crests (or troughs) is one wavelength λ
( = v f ). The spacing between the crests (troughs)
s
received by both observers is also λ = v f , so they
s
detect no change in frequency. http://www.colorado.edu/physics/2000/applets/doppler2.html However, when the car is moving with speed u s towards
one observer and away from the other, they each will
receive a frequency that is different from the source (f s).
The period of the source is Ts = 1f , so if a crest leaves
s
When the car is not moving with respect to observers in
front of and behind the car, the observers receive a
frequency that is the same as that of the siren, i.e., there the source at t1 the next crest will leave the source at t 2,
where ( t 2 − t1 ) = Ts . But during this time interval both
the source and the crest will have moved. is no observed change in frequency. 10 us
s( t1 )
vTs
s( t 2 )
λb λf Ts = t 2 − t1 = 1 f
s
stationary source us Ts In the figure above, s( t1 ) and s( t 2 ) are the positions of
the source at t1 and t 2. So, the first crest will have
moved a distance vTs and the source will have moved a
distance u s Ts when the next crest is emitted by the
source. Since the distance between crests is the
“received” wavelength, we find
λ b = ( v + u s ) Ts (behind the source), and moving source Substituting for Ts we get λ = (v ± us )
,
fs where the plus sign corresponds to λ = λ b and the minus
sign corresponds to λ = λ f . Hence, the received
frequencies are fr = v
v
=
fs.
λ v ± us This expression applies when the receivers are stationary
with respect to the medium and v > u s. λ f = ( v − u s ) Ts (in front of the source), i.e., the source wavelength becomes “stretched”
(“compressed”) behind (in front of) the source. 11 If the source is stationary and the receiver is moving with
speed u r , and approaching the source from the right, we But, the received frequency is f r = 1 T , so
r
fr = v ± ur v ± ur
=
fs.
λ
v have the following scenario. r ( t1 ) and r ( t 2 ) are the
positions of the receiver at times t1 and t 2 when If the receiver approaches the source ( + sign), the successive crests arrive. frequency is higher; if the receiver moves away from the
λa source ( − sign), the frequency is lower. u r Tr moving receiver vTr r ( t1 ) both the receiver and source are moving! From earlier
(v ± us )
we have λ =
fs . λr stationary source r(t 2 ) λ During the time interval between two successive crests,
i.e., t 2 − t1 = Tr , each crest will have moved a distance
vTr and during the same time the receiver will have
moved a distance u r Tr . Then we have
vT ± u T = λ , i.e., T = λ
r rr r We can combine this result with our previous result if ( v ± u r ), where the plus sign means the receiver is moving
towards the source and the minus sign means the
receiver is moving away from the source. ∴ Tr = λ
(v ± us )
=
,
( v ± u r ) ( v ± u r )f s 1 (v ± u r )
f s,
i.e., f r = = Tr ( v ± u s )
where u s and u r are with reference to a stationary
medium. The correct choice of signs is most easily
determined by remembering that the frequency tends to
be higher when the source moves towards the receiver
and the receiver moves towards the source. 12 (a) The person standing by the door is a stationary
observer ahead of an approaching source. Therefore,
we frequency they receive is
v
fr =
fs
(v − us )
Question 3: A police car, with its 300Hz siren blasting,
is moving towards a warehouse at 30m/s intending to = 340
× 300 = 329Hz .
(340 − 30) crash through the door. (a) What frequency does a
person standing by the door hear? (b) What frequency
does the driver of the police car hear reflected from the
warehouse? The speed of sound in still air is 340m/s. (b) So, the frequency reflected from the warehouse is
329Hz. As the police car approaches, this is now the
source frequency; i.e., the police car is a receiver
approaching a stationary source at 30m/s. Therefore,
the frequency the driver receives is
(v + u r )
fr =
fs
v
(340 + 30)
=
× 329 = 358Hz .
340 13 This is similar to the previous problem in that the
approaching car is both a receiver and an emitter. The
frequency received by the moving car approaching a
stationary source is fr = Question 4: A police radar gun uses electromagnetic c+u
fs,
c radiation with a frequency of 2.00GHz. A speeding car where c is the speed of the radar waves and u is the is approaching a stationary policeman using a radar gun. speed of the car. This frequency now becomes the If the policeman measures a change in frequency moving source. The frequency received by the between the signal emitted from the gun and the received
signal is 293Hz, what is the speed of the car? stationary policeman from the approaching car is
c
f r′ =
f s′ ,
c−u (The speed of radar waves is the speed of light, i.e., where f s′ = f r above. c = 3 × 108 m/s.) ∴ f r′ = c (c + u )
(c + u )
fs =
fs
(c − u ) c
(c − u ) (1 + u c ) f = f (1 + u )(1 − u )−1.
=
c
(1 − u c ) s s c
( But u << c , i.e., 1 − u c )−1 ≈ (1 + u c ). 14 ( )( ) ( ) ∴ f r′ ≈ f s 1 + u c 1 + u c ≈ f s 1 + 2 u c ,
2
because u c << 1. () 2u
∴ ∆f = f r′ − f s = f s ,
c c∆f
i.e., u =
2f s “If two or more waves are traveling through a medium,
the resultant disturbance at any point is the algebraic
sum of the individual disturbances.”
Waves that obey this principle are called linear waves. 8 = Principle of superposition 3 × 10 × 293
= 21.98 m/s.
2 × 2 × 109 21.98
× 3600 = 49.1mi/h .)
(21.98 m/s ⇒
1.61 × 103
Strictly speaking, because radar waves are One consequence is that two waves can “pass” through
each other!
y2
y1 y2
y1 + y 2 y1 + y 2 electromagnetic waves (traveling at the speed of light)
we should have used a slightly different form for the y1 + y 2 Doppler shift, i.e., the one appropriate to special
relativity. However, since in this case u ( ) −8
<< 1,
c ≈ 7 × 10 the error is trivially small. y1 y2
y2 y1 y1 (Waves that do not obey this principle are called nonlinear waves.) 15 So, the superposition of several waves of differing
wavelengths and amplitudes produces complex y1 = 6 sin x waveforms. For example, to produce a square wave ... y 2 = 5 sin 2 x
y 3 = 4 sin 3x
y 4 = 3 sin 4 x
y 5 = 2 sin 5x
y 6 = sin 6 x WBX06VD1.MOV A square wave can be expressed as a socalled Fourier
series: f (x) = ∞ 1 nπ sin
x,
L n =1,3,5K n
∑ where L = λ 2 , i.e., onehalf of the wavelength. y = y1 + y 2 + y 3 + y 4 + y 5 + y 6 A Fourier series decomposes any periodic function into a
sum of simple oscillating functions, i.e., sines and/or
cosines. 16 Returning to single frequency (i.e., monochromatic)
waves. Suppose we add two cosine waves. What is the
resultant?
P
r1 S1 r2
S2 For simplicity, the write the phase angles as kr1 = φ1 and kr2 = φ2
∴ y1 + y 2 = A cos(ωt − kr1 ) + A cos(ωt − kr2 )
= A cos(ωt − φ1 ) + A cos(ωt − φ2 ). Using the trignometric relationship:
A + B
A − B
cos A + cos B = 2 cos
cos
,
2
2
we find, Take two separated sources, S1 and S2, with identical
frequency, wavelength and amplitude. Such sources are (φ − φ1 )
(φ + φ1 )
cos ωt − 2
y1 + y 2 = 2 A cos 2
. 2
2 described as coherent. Then at P: Note, it is only this straightforward if the amplitudes of y1 = A cos(ωt − kr1 ) and y 2 = A cos(ωt − kr2 )
Thus, the resultant disturbance at P is: y1 and y 2 are equal. y1 + y 2 = A cos(ωt − kr1 ) + A cos(ωt − kr2 )
How can we work this out ?? () From earlier, (φ2 − φ1 ) = k ( r2 − r1 ) = 2π λ ( r2 − r1 ) ,
where (φ2 − φ1 ) is called the phase difference ( = δ )
between the two waves, and ( r2 − r1 ) is called the path Providing the amplitudes are the same we can use difference ( = ∆r ), i.e., the difference in the length of the trigonometry. two paths. 17 k∆ r (φ + φ1 )
∴ y1 + y 2 = 2 A cos
cos ωt − 2
, 2
2 δ cos ωt − (φ2 + φ1 )
.
or y1 + y 2 = 2 A cos 2
2
We note that the second cosine term is a traveling wave
(φ + φ1 )
with frequency ω and phase angle 2
2 , whereas Thus, if the path difference is an integral number (n) of
wavelengths or the phase difference is an even number
of π , ( y1 + y 2 ) will have a maximum. This condition is
known as constructive interference. the first cosine term “looks” like an amplitude. So we
can write the resultant wave equation as
(φ + φ1 )
y1 + y 2 = A ′ cos ωt − 2
, 2 δ = 2 A cos k∆r .
where the amplitude A ′ = 2 A cos
2 2
The resulting disturbance ( y1 + y 2 ) can only have a
maximum when the amplitude A ′ is a maximum, i.e., (k∆r 2) = nπ () or δ 2 = nπ,
EVEN
i.e., when ∆r = nλ or δ = 2nπ . The resulting disturbance ( y1 + y 2 ) is zero (minimum)
when A ′ = 0 . Then
k∆r = δ = 2 n + 1 π,
2
2
2
ODD )()
∆r = ( n + 12)λ ( i.e., when or δ = (2 n + 1)π. where n = 0, ± 1, ± 2 K .
This condition is known as destructive interference. where n = 0, ± 1, ± 2 K . 18 Let’s look at some specific examples. Take y1 = sin θ and y 2 = sin(θ + δ )
and plot
y1 + y 2 = sin θ + sin(θ + δ ) for various values of the phase difference, δ , between
the waves. Question 5: Two loudspeakers emit 500Hz sound waves
with an amplitude of 0.10mm. If speaker #2 is 1.00m
behind speaker #1, and the two emitted signals are inphase, what is the amplitude of the sound at a point
WBM07AN1.MOV 2.00m in front of speaker #1? (Assume the speed of
sound is 343m/s.) Note that when the phase difference, δ , between the two
waves is 0, 360o ( = 2nπ , with n = 1 ) we have
constructive interference.
When the phase difference, δ , between the two waves is
180o ( = (2 n + 1)π , with n = 0 ) we have destructive interference. 19 #2 #1
x1 • P x2 5.00m
#1 We take the speed of sound as 343m/s. The resultant 2.00 m amplitude is 2.00 m ()
where A = 0.10 mm and k∆x = (2π λ )( x 2 − x1 ).
A ′ = 2 A cos k∆x 2 , We have λ = •P #2 Question 6: Two loudspeakers are 2.00m apart. Both v 343m/s
=
= 0.686 m
f 500Hz emit a 700Hz sound waves that are inphase the source
into a room. A listener stands at the point P, 5.00m in and front of the speakers and 2.00m to one side of the center. ( x 2 − x1 ) = 1.00 m
1.00
= 9.16 rad .
so k∆x = 2π
0.686 ( ) Is the interference at P constructive or destructive or
something between? (Assume the speed of sound is ( ∴ A ′ = 2 A cos k∆x 2 = 2 × 0.10 cos 9.16 2 ) 343m/s.) = 0.0265mm. So, the interference is partially destructive since A ′ < A . 20 #1
2.00 m •P 5.00m
r1 The general wave equation 2.00m r2 Earlier we wrote: #2 y ( t , x ) = A cos(ωt − kx ) ( to describe a traveling wave. However, this is not the ) The amplitude at P is A ′ = 2 A cos k∆r 2 , only expression we could have chosen. For example, where ∆r = r2 − r1 and k = 2π λ . We have y ( t , x ) = B sin(ωt − kx ) and y ( t , x ) = Ce −i(ωt − kx) r1 = (1.00)2 + (5.00)2 = 5.10 m,
r2 = (3.00)2 + (5.00)2 = 5.83m,
and are also traveling waves. They may look different but
they are all solutions of the “general wave equation”
∂2 y 1 ∂2 y
=
.
∂x 2 v 2 ∂t 2 v 341m/s
λ= =
= 0.487 m.
f 700Hz ∴ ∆r = 0.73m.
∆r 0.73
=
= 1.50
In terms of wavelength
λ 0.487 ( = 3 , 2 ) i.e., ∆r = n + 12 λ , with n = 1.
From earlier, this is the condition for totally destructive
interference. Note also, the phase difference is
δ = 2π λ ∆r = 2 ∆r λ π = 3π .
ODD () ( ) Example: if y( t, x) = A cos(ωt − kx) ,
∂y 2
= kA sin(ωt − kx) and ∂ y 2 = − k 2 A cos(ωt − kx) ,
∂x
∂x ∂y ∂ 2y
= −ω 2 A cos(ωt − kx) .
∂t = −ωA sin(ωt − kx) and
∂t 2 k 2 ∂ 2y
∂ 2y 1 ∂ 2y
= =
, i.e.,
,
∂x 2 ω ∂t 2
∂x 2 v 2 ∂t 2
∂ 2y where v is the velocity of the wave. 21 Reflection of waves
Hard boundary.
When a traveling wave or
Incident pulse pulse encounters a change in Soft boundary the medium, part of the wave
(or all of the wave, if it is a However, the situation is quite different for a soft hard medium) is reflected. boundary; there is no change of phase. However, the reflected pulse
or wave is inverted, i.e., it
undergoes a phase change of
Reflected pulse π (180o). WBA03AN4.MOV WBA03AN2.MOV 22 Standing waves We get a standing wave when we pluck a taught string
Consider two waves with the same amplitude and that is clamped at both ends. frequency traveling in opposite directions along the xaxis,
i.e., y R = a cos(ωt − kx ) and y L = a cos(ωt + kx ).
Then applying the superposition principle, the resultant is
y R + y L = a cos(ωt − kx ) + a cos(ωt + kx ) Using the trignometric relationship:
A + B
A − B
,
cos A + cos B = 2 cos
cos
2
2 WBM09AN1.MOV we get
y ( x, t ) = y R + y L = 2a cos(ωt )cos( − kx ) = 2a cos(ωt )cos( kx ).
This is called a standing wave; because the crests and
troughs “stand inplace” but the amplitude varies NODE: a position of zero amplitude.
ANTINODE: a position of maximum amplitude.
The nth harmonic has n antinodes. sinusiodally with a maximum of 2a! See animation ... 23 Here are the first few standing waves on a string fixed at
both ends.
λn n
N
1 N
A
Fundamental, 1st harmonic N
2 3 4 N
A N
A 2nd harmonic
N
N
N
N
A
A
A
3rd harmonic
N
N
N
N
N
A
A
A
A
4th harmonic fn
v
2L 2L
1 1 2L
2 v
2
2L 2L
3 3 2L
4 In the answer to Question 2(d) we stated without proof
T
that the speed of sound (v) in a string is v =
, where T
µ v
4
2L is the tension and µ is the mass per unit length. So, v
2L fn = n v
nT
=
,
2L 2L µ i.e., increasing the tension in a string produces higher
frequency. L 2L
and the
n
v
v
=n
= nf1,
frequencies are given by f n =
λn
2L The wavelengths are given by λ n = where n = 1, 2, 3, K . f1 is called the fundamental
frequency Question 7: A violin string of length 40cm and mass 1.2g
has a frequncy of 500Hz when vibrating in its
fundamental mode. (a) What is the tension in the string,
and (b) where should you place your finger to increase
the frequency to 650Hz? 24 Simple harmonic motion (a) At the fundamental frequency the wavelength λ = 2L = 2 × 0.40 = 0.80 m.
But v = fλ = Simple harmonic motion (SHM) is a very basic kind of ( µ) ⇒ T = (fλ) µ ,
2 T oscillatory motion. The simplest example is a mass
attached to a spring. where µ = m L.
∴ T = (500 × 0.80)2 × 1.2 × 10
0.40 Equilibrium
x −3 = 480Hz . m (b) Since, the velocity of waves on the string is
unchanged, as T and µ are unchanged, in order to
achieve a frequency of 650Hz, the length must be Unstretched
(no mass)
Equilibrium adjusted (to L′). m y ∴ v = f1λ1 = f ′λ ′ , f
i.e., 2f1L = 2f ′L′ ⇒ L′ = 1 L
f′
= 500
× 0.40 = 0.308 m,
650 so you should place your finger 0.308m along the
fingerboard on the violin. m If a mass is displaced from the equilibrium position, in
the absence of frictional forces it will oscillate with
SHM. Let us analyze the motion of the mass on the
vertical spring. 25 If the mass is displaced by y from its equilibrium As before, the maximum displacement A is the position, the restoring force acting on the mass is given amplitude, ω is the by Hookes Law, i.e., angular frequency and
Fy = − ky , δ is the phase angle y( t) when t = 0. In our where k is the spring constant.
A By Newton’s 2nd Law
Equilibrium
y m ∴ d2y
k
= − y.
m
dt 2 This expression defines SHM, i.e.,
if the acceleration of an object is proportional to its
displacement and is oppositely directed, the object
will move with SHM.
A solution to the differential equation is
y ( t ) = A cos(ωt + δ ), where A, ω and δ are constants. case, if the mass is
pulled down a distance ma y = Fy,
d2y
i.e., m 2 = − ky .
dt t A and released, then δ = 0 , so y ( t ) = A cos ωt . Then
dy
= −ωA sin ωt ( = v )
dt
and
d2y
2
2
= a ).
2 = −ω A cos ωt = −ω y (
dt Thus, y ( t ) = A cos ωt is a solution provided
k
ω2 = .
m
The frequency of the oscillations is f = ω 2π and the
periodic time (for one complete oscillation) is T = 1f . 26 ∴T = 1k
.
2π m The solution is y ( t ) = A cos(ωt + δ ). The mass starts
oscillating at t = 0 when y = 8cm, so δ = 0 . Note that the period of SHM is independent of the
amplitude. (In music, it means that the pitch or (a) The speed of the mass is given by v = ωA sin ωt , so frequency of a note struck on any stringed instrument, the maximum speed is for example, is independent of how loudly the note is
played.)
We would obtain the same type of solution for a
horizontal spring, viz:
x ( t ) = A cos(ωt + δ ). v max = ωA = A k
12
= 0.08 ×
= 0.438m/s.
m
0.4 (b) When y ( t ) = 4cm = 0.04 m, we have
1
π
0.04 = 0.08 cos ωt ⇒ ωt = cos −1 = . 2 3
π
3
∴ v = ωA sin ωt = 0.438 × sin = 0.438 × 3
2 = 0.379 m/s. Question 8: A 0.40kg mass attached to a spring of force
constant 12N/m oscillates with an amplitude of 8cm.
Find (a) the maximum speed of the mass, (b) the speed
and acceleration of the mass when it is at y = 4cm from
the equilibrium position, and (c) the time it takes the
mass to travel from y = 0 to y = 4cm. Also, we have
π
a = −ω 2 A cos ωt = v maxω cos 3
= 0.438 × 12 1
× = 1.20 m/s2 .
0.4 2 27 Physical pendulum
Any rigid object that is free to rotate will oscillate about
its equilibrium position when displaced and released. We
show now that when such an object is released, its
motion will be SHM. Using
(c) When y = 0 , Newton’s 2nd Law for π
A cos ωt = 0 ⇒ ωt = cos −1 0 = .
2
π
From (b), when y = 4cm, ωt ′ = .
3
∴ ∆t = t ′ − t = rotation θD
cm
r
Mg rr
Iα = τ , we have
d 2θ
I 2 = −(D sin θ)Mg .
dt 1 π π
π π 0.4
=
=
−
ω 2 3 6ω 6 12 = 0.096s ( = 96ms). If θ is small, sin θ → θ , then I i.e., d 2θ
= −MgDθ,
dt 2 d 2θ
MgD
=−
θ.
I
dt 2 Note that this equation satisfies our definition of SHM.
A suitable general solution is
θ = A cos(ωt + δ ). 28 ∴ dθ
= −ωθ max sin ωt
dt d 2θ
and 2 = −ω 2θ max cos ωt = −ω 2θ .
dt Thus, if θ = θ max cos ωt is a solution, MgD
MgD
ω2 =
, i.e., ω =
.
I
I
But the swing frequency is f = ω 2π and the period for
one complete oscillation is T = 1f .
2π
I
∴T =
= 2π
.
ω
MgD Question 9: A student measures the length of his leg,
from hip to heel, to be 0.90m. (a) What is the frequency
of the pendulum motion of the student’s leg? (b) What is
the period?
(Model the leg as a rod of uniform crosssection and
pivoted at one end, i.e., the hip.) Note that providing θ is small the periodic time is
independent of the amplitude. 29 We have T = 1
I
1 MgD
= 2π
⇒f =
.
f
MgD
2π
I (a) The moment of inertia of a uniform rod pivoted at
1
one end is I = ML2. Since the leg is assumed uniform,
3
1
the center of mass is at a distance D = L from the
2 Simple pendulum
The simple pendulum is a special case of the physical (or
compound) pendulum. In this case, if we ignore the
mass of the string, the moment of inertia of the bob of
mass m about the pivot point is
θ pivot. ∴f = 1 MgD 1
=
2π
I
2π ()
() Mg L 2
1 3g
2 = 2 π 2L
1 ML
3 r
T 1
1
= 1.55s.
(b) T = =
f 0.64 and the distance of the center s
r
Ft r
mg 1 3 9.81
=
×
= 0.644Hz .
2π 2 0.90 I = ml2, l θ m
r
Fr of mass of the bob fom the
pivot point is D = l.
The solution is again
θ = θ max cos ωt , but with ω = mgD
=
I mgl
=
ml2 g
. The period for one
l pendulum motion, i.e., ~ 0.8s. For a normal walking entire oscillation is T = 1f = 2π ω , i.e.,
l
T = 2π .
g pace, one stride in just under a second is about right. Note that we can write the arc length as s = lθ , so that When you walk, swinging your free leg forward to take
another stride corresponds to half a period of this 30 s = lθ max cos ωt .
• The tangential speed of the bob is
ds
v t = = −ωlθ max sin ωt ,
dt
which is zero when ωt = nπ , i.e., when s = smax = lθ max .
The tangential speed is a maximum when sin ωt = 1, i.e.,
2n + 1
π . That occurs when s = 0.
when ωt =
2
• The tangential accceleration of the bob is
d 2s
a t = 2 = −ω 2lθ max cos ωt ,
dt
2n + 1
π , i.e., when s = 0. The
which is zero when ωt =
2 Question 10: The pendulum shown below is an
“arrested” pendulum. There is a peg P below the pivot
point O so when the pendulum swings to the right, the
string makes contact with the peg, which causes the bob
to swing on a different arc. If the length of the string is
78.0cm, and the distance OP is onehalf the length of the
string, what is the periodic time of the pendulum?
O• •P tangential acceleration is a maximum when sin ωt = 1,
i.e., when ωt = nπ . That occurs when s = smax = lθ max . 31 We can think of the pendulum as two halves ... when it
swings from the left, the onequarter period is
O•
1
l
T1 = ⋅ 2π
4
g
l •P and when it swings past the
l Question 11: We have just seen that the maximum speed vertical, the onequarter period is of the bob of a pendulum is ωlθ max . The maximum 2 1
T2 = ⋅ 2π
4 l 2.
g mechanical energy. (a) Show that the result is the same.
(b) The bob of a pendulum is released at an angle of 25o Thus, the total period is
π l π l 1 l 2( T1 + T2 ) = 2
+ = π 1 + 2 g 2 2g 2g = 1.71π speed can also be found using the conservation of l
0.78
= 1.71π
= 1.51s.
g
9.81 from the vertical; if the length of the string is 35cm,
calculate the maximum speed of the bob at the bottom of
the swing using both methods. The behavior of the arrested pendulum was discussed by
Galileo ... can you guess why?? 32 (a) Using the conservation of mechanical energy, at the (b) Using the conservation of mechanical energy, at the bottom of the swing, where the speed is a maximum
12
mv = mgh = mgl(1 − cos θ),
2
θ bottom of the swing, where the speed is a maximum we l l cos θ i.e., v = 2 gl(1 − cos θ) ,
m h where θ = θ max . But cos θ = 1 − sin2 θ . find v = 2 gl(1 − cos θ)
= 2 × 9.81 × 0.35 1 − cos 25o = 0.802 m/s. ( Using the SHM expression v = ωlθ = lθ So, when θ is small, cos θ → 1 − θ2 .
1 ∴ v = 2 gl1 − 1 − θ2 2 . ( ) ) = g
= θ gl
l 25 × 2π
9.81 × 0.35 = 0.809 m/s.
360 1 1
But 1 − θ2 2 ≈ 1 − θ2,
2 ( so ) 1
v = 2 gl θ2 = ( gl )θ = ωlθ ,
2 since ω = g l , i.e., v max = ωlθ max . DISCUSSION PROBLEM [14.3]:
Note that these results are slightly different; which one is
the more reliable? Why? 33 ...
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 Spring '08
 Guzman
 Physics

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