In Brazil, a country that was undergoing a rapid inflation before 1994, many transactions were conducted in dollars rather than in reals, the domestic currency. During this period, the U.S.. dollar served what property or properties in Brazil?
Mastering Physics
Friday, December 21, 2012
HW1Q11
HW1Q10
HW1Q1
Thursday, November 29, 2012
What Is a Sound Wave?
Part A = Propagation of pressure fluctuations in a medium
Part B = Pressure fluctuations travel along the direction of propagation of the sound wave.
Part C = yes
Part D = BC
Part E = 1000,0.344
Part F = It is perceived as louder.
Part G = It is perceived as higher in pitch.
Part H = the harmonic content
Solutions Below:
Part E
Part B = Pressure fluctuations travel along the direction of propagation of the sound wave.
Part C = yes
Part D = BC
Part E = 1000,0.344
Part F = It is perceived as louder.
Part G = It is perceived as higher in pitch.
Part H = the harmonic content
Solutions Below:
Learning Goal: To understand the nature of a sound wave, including its properties: frequency, wavelength, loudness, pitch, and timbre.
Sound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to understand the physical nature of a sound wave and its properties to correct common misconceptions about sound propagation.
Most generally, a sound wave is a longitudinal wave that propagates in a medium (i.e., air). The particles in the medium oscillate back and forth along the direction of motion of the wave. This displacement of the particles generates a sequence of compressions and rarefactions of the medium. Thus, a sound wave can also be described in terms of pressure variations that travel through the medium. The pressure fluctuates at the same frequency with which the particles’ positions oscillate.
When the human ear perceives sound, it recognizes a series of pressure fluctuations rather than displacements of individual air particles.
Part ASound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to understand the physical nature of a sound wave and its properties to correct common misconceptions about sound propagation.
Most generally, a sound wave is a longitudinal wave that propagates in a medium (i.e., air). The particles in the medium oscillate back and forth along the direction of motion of the wave. This displacement of the particles generates a sequence of compressions and rarefactions of the medium. Thus, a sound wave can also be described in terms of pressure variations that travel through the medium. The pressure fluctuates at the same frequency with which the particles’ positions oscillate.
When the human ear perceives sound, it recognizes a series of pressure fluctuations rather than displacements of individual air particles.
Based on the information presented in the introduction of this problem, what is a sound wave?
A. Propagation of sound particles that are different from the particles that comprise the medium
B. Propagation of energy that does not require a medium
C. Propagation of pressure fluctuations in a medium
D. Propagation of energy that passes through empty spaces between the particles that comprise the medium
C. Propagation of pressure fluctuations in a medium
Part BA. Propagation of sound particles that are different from the particles that comprise the medium
B. Propagation of energy that does not require a medium
C. Propagation of pressure fluctuations in a medium
D. Propagation of energy that passes through empty spaces between the particles that comprise the medium
C. Propagation of pressure fluctuations in a medium
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?
A. The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.
B. Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.
C. Pressure fluctuations travel along the direction of propagation of the sound wave.
C. Pressure fluctuations travel along the direction of propagation of the sound wave.
Part CA. The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.
B. Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.
C. Pressure fluctuations travel along the direction of propagation of the sound wave.
C. Pressure fluctuations travel along the direction of propagation of the sound wave.
Does air play a role in the propagation of the human voice from one end of a lecture hall to the other?
A. yes
B. no
A. yes
Part DA. yes
B. no
A. yes
The graphs shown here represent pressure variation versus time recorded by a microphone. Which could correspond to a sound wave?
Enter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think all three graphs could represent sound waves, enter ABC.
BC
Enter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think all three graphs could represent sound waves, enter ABC.
BC
Part E
The next graph shows a sound wave consisting of a sinusoidal displacement of air particles versus time, as recorded at a fixed location. For sinusoidal waves, it is possible to identify a specific frequency (rate of oscillation) and wavelength (distance in space corresponding to one complete cycle).
Taking the speed of sound in air to be 344 m/s, what are the frequency f and the wavelength lambda of the sound wave shown in the graph?
Express your answers in, respectively, hertz and meters to three significant figures. Separate the two answers with a comma.
The velocity of a wave equals its frequency times the wavelength:
v = fλ
V is given as 344m/s and T is shown by the graph to be 10-3s = 0.001s. Since f = 1/T, f = 1/0.001 = 1000Hz
Therefore:
v = fλ
344 = 1000λ
λ = 0.344
So the answer is:
1000,0.344
Note: pay attention to the labellings of the axes. For this problem, the x-axis was labeled as t(s), so it’s representing period, not λ
Part FTaking the speed of sound in air to be 344 m/s, what are the frequency f and the wavelength lambda of the sound wave shown in the graph?
Express your answers in, respectively, hertz and meters to three significant figures. Separate the two answers with a comma.
The velocity of a wave equals its frequency times the wavelength:
v = fλ
V is given as 344m/s and T is shown by the graph to be 10-3s = 0.001s. Since f = 1/T, f = 1/0.001 = 1000Hz
Therefore:
v = fλ
344 = 1000λ
λ = 0.344
So the answer is:
1000,0.344
Note: pay attention to the labellings of the axes. For this problem, the x-axis was labeled as t(s), so it’s representing period, not λ
A certain sound is recorded by a microphone. The same microphone then detects a second sound, which is identical to the first one except that the amplitude of the pressure fluctuations is larger. In addition to the larger amplitude, what distinguishes the second sound from the first one?
A. It is perceived as higher in pitch.
B. It is perceived as louder.
C. It has a higher frequency.
D. It has a longer wavelength.
B. It is perceived as louder.
Part GA. It is perceived as higher in pitch.
B. It is perceived as louder.
C. It has a higher frequency.
D. It has a longer wavelength.
B. It is perceived as louder.
A certain sound is recorded by a microphone. A second sound that has twice the frequency is detected by the same microphone. In addition to the higher frequency, what distinguishes the second sound from the first one?
A. It is perceived as higher in pitch.
B. It is perceived as louder.
C. It has a higher amplitude.
D. It has a longer wavelength.
A. It is perceived as higher in pitch.
Part HA. It is perceived as higher in pitch.
B. It is perceived as louder.
C. It has a higher amplitude.
D. It has a longer wavelength.
A. It is perceived as higher in pitch.
What varies between two tones that are different in timbre, that is, two tones that have the same fundamental frequency but are produced, say, by different musical instruments?
Note that Figures (b) and (c) from Part D could represent tones with different timbre.
A. the pitch
B. the harmonic content
C. nothing
B. the harmonic content
Note that Figures (b) and (c) from Part D could represent tones with different timbre.
A. the pitch
B. the harmonic content
C. nothing
B. the harmonic content
Energy of Harmonic Oscillators
Energy of Harmonic Oscillators
Part A = A
Part B = A
Part C = moving toward equilibrium.
Part D = C
Part E = C
Part F = D
Part G = 3/8kA2
Solutions Below:
Part A
A
B
C
D
The maximum PE is when the spring is fully compressed. D might look like the right answer, but actually A is when the spring is at amplitude. Even though it’s stretched, that still counts as compression:
A
Part B
A
B
C
D
When PE is maximized, KE is minimized, so the correct answer is the same as from Part A:
A
Part C
A. at the equilibrium position.
B. at the amplitude displacement.
C. moving to the right.
D. moving to the left.
E. moving away from equilibrium.
F. moving toward equilibrium.
KE is maximized when PE is minimized. PE is greatest at amplitude and KE is greatest at equilibrium. So choice F is correct:
F. moving toward equilibrium.
Part D
A
B
C
D
KE is greatest at equilibrium:
C
Part E
A
B
C
D
Minimum PE is when KE is greatest, i.e. at equilibrium:
C
Part F
When U = KE, U = 1/2Umax (this is just a fact, which we won’t bother solving for here). So:
U = 1/2(U)
1/2kx2 = 1/2(1/2kA2)
x2 = 1/2A2
x = sqrt(1/2A2)
x = Asqrt(2)/2
Note- if it isn’t obvious where the sqrt(2)/2 came from, use 2/4 for the fraction above instead of 1/2:
x = sqrt((2/4)A2)
x = Asqrt(2)/2
The diagram doesn’t give this as an answer, but it does give -Asqrt(2)/2, which is equivalent:
D
Part G
Since total energy = KE + PE and we only have enough information to find PE, we can work backwards by first finding the maximum PE (the PE when the spring is fully compressed) and then subtracting the PE at point B.
Maximum PE:
PEmax = 1/2kA2
PE at Point B:
PEB = 1/2k(A/2)2
PEB = 1/2k(A2/4)
PEB = 1/8k(A2)
Now find the difference between the PEs, and that difference must be KE since there is no friction:
KEB = PEmax – PEB
KEB = 1/2kA2 – 1/8k(A2)
KEB = 3/8kA2
Part A = A
Part B = A
Part C = moving toward equilibrium.
Part D = C
Part E = C
Part F = D
Part G = 3/8kA2
Solutions Below:
Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure . Assume that the force constant k, the mass of the block, m, and the amplitude of vibrations, A, are given. Answer the following questions:
Part A
Which moment corresponds to the maximum potential energy of the system?
A
B
C
D
The maximum PE is when the spring is fully compressed. D might look like the right answer, but actually A is when the spring is at amplitude. Even though it’s stretched, that still counts as compression:
A
Part B
Which moment corresponds to the minimum kinetic energy of the system?
A
B
C
D
When PE is maximized, KE is minimized, so the correct answer is the same as from Part A:
A
Part C
Consider the block in the process of oscillating. If the kinetic energy of the block is increasing, the block must be:
A. at the equilibrium position.
B. at the amplitude displacement.
C. moving to the right.
D. moving to the left.
E. moving away from equilibrium.
F. moving toward equilibrium.
KE is maximized when PE is minimized. PE is greatest at amplitude and KE is greatest at equilibrium. So choice F is correct:
F. moving toward equilibrium.
Part D
Which moment corresponds to the maximum kinetic energy of the system?
A
B
C
D
KE is greatest at equilibrium:
C
Part E
Which moment corresponds to the minimum potential energy of the system?
A
B
C
D
Minimum PE is when KE is greatest, i.e. at equilibrium:
C
Part F
At which moment is K = U?
When U = KE, U = 1/2Umax (this is just a fact, which we won’t bother solving for here). So:
U = 1/2(U)
1/2kx2 = 1/2(1/2kA2)
x2 = 1/2A2
x = sqrt(1/2A2)
x = Asqrt(2)/2
Note- if it isn’t obvious where the sqrt(2)/2 came from, use 2/4 for the fraction above instead of 1/2:
x = sqrt((2/4)A2)
x = Asqrt(2)/2
The diagram doesn’t give this as an answer, but it does give -Asqrt(2)/2, which is equivalent:
D
Part G
Find the kinetic energy K of the block at the moment labeled B.
Express your answer in terms of k and A.
Express your answer in terms of k and A.
Since total energy = KE + PE and we only have enough information to find PE, we can work backwards by first finding the maximum PE (the PE when the spring is fully compressed) and then subtracting the PE at point B.
Maximum PE:
PEmax = 1/2kA2
PE at Point B:
PEB = 1/2k(A/2)2
PEB = 1/2k(A2/4)
PEB = 1/8k(A2)
Now find the difference between the PEs, and that difference must be KE since there is no friction:
KEB = PEmax – PEB
KEB = 1/2kA2 – 1/8k(A2)
KEB = 3/8kA2
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