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```Physics 2149 Review Set 3
1) A load on a spring vibrates in simple harmonic motion with an amplitude of 4.00 cm and a frequency of
0.500 Hz. (a) What are the maximum magnitudes of the displacement, velocity and acceleration? (b)
Determine the displacement, velocity, and acceleration of the load 1.50 s after if it passes its equilibrium
position moving in the positive direction.
amax = Aω 2
v = Aω
max
= A ( 2π f )
= A2π f
smax = A
= ( 4.00 cm )( 2π ) ( 0.500 s −1 )
= 4.00 cm
= 12.6 cm / s
s = A sin ωt
(
= ( 4.00 cm ) sin 2π ( 0.5 s −1 ) (1.5 s )
2
= ( 4.00 cm )( 2π ) ( 0.500 s −1 )
2
2
= 39.4 cm / s 2
v = Aω cos ωt
)
(
= (12.6 cm / s ) cos 2π ( 0.5 s −1 ) (1.5 s )
= −4.00 cm
)
= 0 cm / s
a = − Aω sin ωt
2
(
= − ( 39.4 cm / s 2 ) sin 2π ( 0.5 s −1 ) (1.5 s )
)
= 39.4 cm / s 2
2) A 4.00 kg load is suspended from the end of an elastic cable that has negligible mass. When the load is
displaced vertically through 2.50 cm and released it vibrates in simple harmonic motion at a frequency
of 1.25 Hz. Determine (a) The period of the vibration, (b) the spring constant of the cable, (c) the
maximum vibratory speed of the load, (d) the maximum vibratory acceleration of the load, and (e) the
acceleration of the load when its displacement is 1.25 cm.
f =
1
1
= 0.8 s
a) T = =
f 1.25 Hz
k
⇒
m
vmax = Aω
b) k = ( 2π f ) m
2
c)
= ( 2π (1.25) ) ( 4 kg ) = 246 kg / s 2
2
amax = Aω 2
= (2.5 cm) ( 2π (1.25 s -1 ) )
=19.6 cm/s
a = −ω 2 s
= (2.5 cm) ( 2π (1.25 s -1 ) )
d)
1
2π
= 154 cm / s 2
2
e)
= - ( 2π (1.25 s -1 ) ) (1.25 cm)
2
=-77.0 cm/s 2
3) A 2.00 kg mass oscillates in SHM with an amplitude of 4.00 cm and a frequency of 7.96 Hz.
a) Find the spring constant.
b) Find the acceleration when the displacement is 2.00 cm.
c) The total energy of the system.
d) The magnitude of the velocity when the displacement is 2.50 cm
1 k
⇒
f =
a = −ω 2 s
2π m
a) k = ( 2π f ) m
2
(
= 2π ( 7.96 s −1 )
b)
) ( 2.00 kg )
2
= 5.00 × 103 kg / s 2
(
= − 2π ( 7.96 s −1 )
= −50.0 m / s 2
) ( 0.02 m )
2
1 2
kA
2
1
2
= ( 5.00 × 103 kg / s 2 ) ( 0.04 m )
2
= 4.00 J
v=
E=
c)
k 2 2
(A − s )
m
5.00 × 103 kg / s 2
=
2.00 kg
d)
((0.04 m )
2
− (.025 m )
2
)
= 1.56 m / s
4) Show that for a simple pendulum oscillating at small angle θ the
natural frequency is given by
1 g
f =
2π l
FR = −mg sin θ
where we have used the fact that sin θ = θ for small θ,
= − mgθ
mg
=−
s
l
s
and θ = .
l
Comparing this to the restoring force for the spring: FR = − ks we can make the identification:
mg
k=
l
We now substitute this result into our equation for the natural frequency of the spring.
1 k
f =
2π m
=
1
2π
mg / l
QED
m
1 g
2π l
5) A load on a vibrating spring vibrates in harmonic motion at a natural frequency of 10.0 Hz and initial
amplitude of 5.00 cm. After 5 oscillations the amplitude has reduced to 1/2 its original value. What is
the amplitude at 0.25 s.
First we must solve of γ.
The time for 5 oscillations is:
A(t ) = A0 e −γ t / 2 ⇒
=
N
⇒
t
N 5
t = = = 0.5 s
f 10
f =
1
− γ 0.5 s / 2
A0 = A0 e ( ) ⇒
2
⎛1⎞
ln ⎜ ⎟ = ( −0.25 s ) γ ⇒
⎝2⎠
⎛1⎞
γ = − ln ⎜ ⎟ / ( 0.25 s )
⎝2⎠
= 2.77 s −1
A(t ) = A0 e −γ t / 2 ⇒
= ( 5.00 cm ) e
= 3.54 cm
(
)
− 2.77 s −1 ( 0.25 s ) / 2
⇒
6) A load on a vibrating spring vibrates in harmonic motion at a natural frequency of 2.50 Hz and initial
amplitude of 20.0 cm. After 5 oscillations the amplitude has reduced to 1/2 its original value. What is
the displacement, s, at 0.50 seconds.
A(t ) = A0 e −γ t / 2 ⇒
1
−γ 2 s / 2
A0 = A0 e ( ) ⇒
2
⎛1⎞
ln ⎜ ⎟ = ( −1 s ) γ ⇒
⎝2⎠
N
f = ⇒
t
5
N
t= =
= 2.0 s
f 2.5
⎛1⎞
γ = − ln ⎜ ⎟ / (1 s )
⎝2⎠
= 0.693 s −1
s (t ) = A0 e−γ t / 2 sin (ωt ) ⇒
(
s (0.50) = (16.8 cm ) sin 2π ( 2.50 s −1 ) ( 0.5s )
A(t ) = A0 e −γ t / 2 ⇒
A(0.50 s ) = ( 20.0 cm ) e
(
)
− 0.693 s −1 ( 0.50 s ) / 2
= 16.8 cm
)
= 16.8 cm
7) A sound wave is sent into the ground. On the surface layer the sound travels at 500 m/s, in the layer
below the sound travels at 1500 m/s. What is the critical angle of incidence for the sound wave where
the angle of refraction in the second layer will be 90°?
sin α1 v1
= ⇒
sin α 2 v2
sin α1 =
v1
sin α 2
v2
500 m / s
sin 90° ⇒
1500 m / s
1
α1 = sin −1 = 19.5°
3
8) A small sound source emits waves uniformly in all directions with a power of 4.00 mW and at a
frequency of 1000 Hz. The speed of sound in air is 331 m/s and the density of air is 1.29 kg/m3.
Determine (a) the sound intensity at a distance of 10.0 m from the source, (b) the amplitude of the wave,
and (c) the intensity at a distance of 20.0 m from the source. Assume no energy losses.
=
2
P
I=
A'
P
=
4π r 2
4.00 × 10−3 W
=
2
4π (10.0 m )
I1 ⎛ r2 ⎞
=⎜ ⎟ ⇒
I 2 ⎝ r1 ⎠
I = 2π f A ρ v ⇒
2
A=
=
2
2
I
2π f 2 ρ v
2
2
3.18 × 10−6 W / m 2
2π 2 (1000 Hz ) (1.29 kg / m3 ) ( 331 m / s )
⎛r ⎞
I 2 = ⎜ 1 ⎟ I1
⎝ r2 ⎠
2
⎛ 10 ⎞
= ⎜ ⎟ ( 3.18 × 10−6 W / m 2 )
⎝ 20 ⎠
−8
= 3.18 × 10−6 W / m 2 = 1.94 × 10 m
= 7.95 × 10−7 W / m 2
9) A small sound source radiates uniformly in all directions. If the sound intensity level is 50.0 dB at a
distance of 1.00 m from the source, at what distance is the intensity level 30.0 dB?
2
⎛I ⎞
⎛I ⎞
LI 2 − LI1 = 10 log ⎜ 2 ⎟ − 10 log ⎜ 1 ⎟
⎝ I0 ⎠
⎝ I0 ⎠
⎛I ⎞
= 10 log ⎜ 2 ⎟
⎝ I1 ⎠
2
⎛r ⎞
= 10 log ⎜ 1 ⎟ ⇒
⎝ r2 ⎠
⎛ r ⎞
50 − 30 = 20 log ⎜ 1 ⎟ ⇒
⎝ 1.00 ⎠
20
⎛ r ⎞
= log ⎜ 1 ⎟ ⇒
20
⎝ 1.00 ⎠
r
101 = 1 ⇒
1.00
r1 = 10 m
10) A sound wave with a wavelength of 75.0 cm and amplitude of 12.0 mm travels at speed of 330 m/s from
left to right. If the displacement at the origin is initially zero and the particle is moving downward (a)
write the equation of the wave, and (b) determine the displacement at a location x = 95.4 cm after an
elapsed time of 1.00 ms.
v=λf ⇒
⎡
⎛ x
⎞⎤
s ( t , x ) = (12.0 mm ) sin ⎢ 2π ( 440 Hz ) ⎜
− t ⎟⎥
v 330 m / s
= 440 Hz
f = =
⎝ 330 ⎠ ⎦
⎣
λ
0.75 m
⎡
⎛ 0.954
⎞⎤
− 1.00 ms ⎟ ⎥
s (1.00 ms, 0.954 m ) = (12.0 mm ) sin ⎢ 2π ( 440 Hz ) ⎜
⎝ 330
⎠⎦
⎣
= (12.0 mm ) sin ⎡⎣ 2π ( 440 Hz )( 2.89 ms − 1.00 ms ) ⎤⎦
= (12.0 mm ) sin ( 300° )
= −10.4 mm
11) A 2.80 m long metal bar is clamped at a position 70.0 cm from one end. If the speed of sound in the bar
is 5470 m/s, determine (a) the fundamental frequency, (b) the frequency of the third harmonic, and (c)
the frequency of the second overtone, (d) draw the fundamental frequency and the first overtone.
nv
For vibrating rods: f n =
so
L
1( 5470 m / s )
3 ( 5470 m / s )
5 ( 5470 m / s )
f1 =
= 1950 Hz , f3 =
= 5860 Hz , f5 =
= 9770 Hz
2.80 m
2.80 m
2.80 m
0.70 m
2.10 m
•
12) A stationary cop shoots his radar gun with a frequency of 10.00 GHz at an approaching car. If the speed
of the radar is 3.00×108 m/s and the radar gun records the reflected wave at a frequency of 12.22 GHz,
what is the speed of the car?
Cop
c−v
c −u
c − ( −vcar )
= fs
c−0
c + vcar
= fs
c
f car = f s
x
Car
Now look at the wave reflected back off the car.
Car
c−v
c−u
c−0
= f car
c − vcar
f = fs
x
Cop
c
c − vcar
Note that in this picture the velocity of the source is u = vcar , and the velocity of the receiver is v = 0 .
Combine the two results:
c
f = f car
c − vcar
= f car
= fs
c + vcar
c
c c − vcar
= fs
c + vcar
⇒
c − vcar
fc − fvcar = f s c + f s vcar ⇒
vcar = c
(f
(f
− fS )
(12.22 − 10.00 ) = 3.00 ×107 m / s
= ( 3.00 × 108 m / s )
+ fS )
(12.22 + 10.00 )
```