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course phy122
Your solution, attempt at solution:If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
010. `query 9
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Question: Query introductory set 6, problems 1-10
explain how we know that the velocity of a periodic wave is equal to the product of its wavelength and frequency
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Your Solution:
#1: If a wave has the characteristic that 40 equally spaced peaks pass a given point every second, with the wave traveling at 160 m/s then what is the distance between peaks?
Solution:
40 peaks in one second
Since the wave travels at 160 m/s, the section of the wave that passes in a second must be 160 m long.
There are 40 peaks in 160 meters
I will do the following: 160/40= 4 m/peak
#2: what is the wavelength of a wave if careful observation shows that its frequency is 50 cycles/sec, and that the wave travels at 200m/sec?
Solution:
50peaks in one second, section the wave passes in one sec must have 50 peaks
Travels at 200 m/s, the section of the wave that passes in a single second must be 200 m long.
Distance between peaks= 200m/50 peaks= 4 m/peak
#3: What is the velocity of a wave propagation if the wave has a frequency of 25 cycles/sec, and if the wavelength is 2 m?
Solution:
The section of wave which passes in one second contains 25 peaks. Every peak is 2 meters from the next peak. This 1 sec section of the wave has 25 peaks.
2m x 25 peaks= 50 m long
The wave moves 50 meters in one second, velocity will therefore be 50m/sec
#4: Find the frequency of a wave with whose wavelength is 5 m and which propogates at the velocity of 30m/sec
Solution:
I will visualize a one second section of the wave, it is 30 meters long, with peaks 5 m apart. This one second section therefore contains 30 meters/ (5m/peak)=6 peaks
Therefore in one second, 6 peaks will pass.
The frequency will therefore be 6 peaks/ second.
#5: If a wave with 3 meters between peaks passes at 9 meters/sec, what is the period of the wave?
Solution:
A one second section of the wave is 9 m long, and has peaks 3 m apart. Each peak therefore spans 3/9 of the 1 sec second. The time between peaks is 3/9=0.3333sec
#6 What is the velocity of a wave whose wavelength and frequency are 5 m and 50 Hz, respectively?
Solution:
Our picture of a 1 sec section of the wave will show 50 peaks with 5 m between peaks. This section will therefore be 50 x 5m=250 m long and the wave will travel at 250 m/s.
A one second section of the wave is 250 m long and has peaks, so the wave must be moving at 250 m/s.
#7: What is the velocity of a wave if its peak to peak cycles each require .000077 sec and its wavelength is .196 neters?
Solution: A single cycle, from peak to peak is .196 meters long and passes in .000077 sec. The wave therefore travels .196 meters in a time interaval of .0000077 sec so:
.196m/.000077sec=2545 m/sec
#8: A wave whose cycles have period .000052 sec travels a velocity 12900 m/sec. What is it’s wavelength?
Solution:
If a wave is traveling at 12900m/s, then in .000052 seconds the wave will travel through a peak to peak cycle. During .000052 second, the wave travels (12900m/s)(.000052sec)=.671m
The wavelength is .671 m.
# 9 and 10 are to completed by university physics students.
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Given Solution:
** we know how many wavelength segments will pass every second, and we know the length of each, so that multiplying the two gives us the velocity with which they must be passing **
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Question: explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity
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Your Solution:
If we know how far it is between peaks and how fast the wavetrain is passing, we can dicide the distance between the peaks by the velocity to see how much time passes between peaks at a given point. So period=wavelength /velocity.
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Given Solution:
** If we know how far it is between peaks (wavelength) and how fast the wavetrain is passing (velocity) we can divide the distance between peaks by the velocity to see how much time passes between peaks at a given point. That is, period is wavelength / velocity. **
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Question: explain why the equation of motion at a position x along a sinusoidal wave is A sin( `omega (t - x / v) ) if the equation of motion at the x = 0 position is A sin(`omega t)
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Your Solution:
What happens at x at clock time t is what happened at the origin at a certain earlier clock time.
The time required to for a pulse to propagate from the origin to position x is x / v.
So What happens at x at clock time t is what happened at the origin at clock time t - x / v.
At clock time t - x / v the point at the origin was at y position y = A sin (omega (t - x / v)).
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Given Solution:
** the key is the time delay. Time for the disturbance to get from x = 0 to position x is x / v. What happens at the new position is delayed by time x/v, so what happens there at clock time t happened at x=0 when clock time was t = x/v.
In more detail: If x is the distance down the wave then x / v is the time it takes the wave to travel that distance. What happens at time t at position x is what happened at time t - x/v at position x=0.
That expression should be y = sin(`omega * (t - x / v)).
The sine function goes from -1 to 0 to 1 to 0 to -1 to 0 to 1 to 0 ..., one cycle after another. In harmonic waves the motion of a point on the wave (think of the motion of a black mark on a white rope with vertical pulses traveling down the rope) will go thru this sort of motion (down, middle, up, middle, down, etc.) as repeated pulses pass.
If I'm creating the pulses at my end, and that black mark is some distance x down in rope, then what you see at the black mark is what I did at time x/v earlier. **
STUDENT COMMENT (University Physics):
According to the Y&F book (p.553) we get the expression for a sinusoidal wave moving the the +x-direction with the equation:
Y(x,t) = A*cos[omega*(t-x/v)]
I am not sure where the sine came from in the equation in the question. The book uses the cosine function to represent the waves motion.
The choice of the cosine function is arbitrary. Either function, or a combination of both, can come out of the solution to the wave equation (that's the partial differential equation which relates the second derivative with respect to position to the second derivative with respect to time).
The sine and cosine functions differ only by a phase difference of 90 degrees, and either can be used to describe simple harmonic motion or the motion of harmonic waves. The choice simply depends on the initial conditions of the system.
We don't want to get into solving the wave equation here, but the point can be illustrated by considering simple harmonic motion, which is characterized by F_net = - k x (leading to m x '' = - k x or x '' = -k/m * x, where derivatives are with respect to time).
The general solution to the equation x '' = - k / m * x is x = B sin(omega t) + C cos(omega t), where B and C are arbitrary constants and omega = sqrt(k/m).
B sin(omega t) + C cos(omega t) = A sin(omega t + phi), where A and phi are determined by B, C and the choice to use the sine function on the right-hand side.
B sin(omega t) + C cos(omega t) = A cos(omega t + phi), where A and phi are determined by B, C and the choice to use the cosine function on the right-hand side. The value of A will be the same as if we had used the sine function on the right, and the value of phi will differ by 90 degrees or pi/2 radians.
STUDENT QUESTION
I don’t understand why we would subtract x/v from the omega*t value. Why wouldn’t it be addition?
INSTRUCTOR RESPONSE
What happens at x at clock time t is what happened at the origin at a certain earlier clock time.
The time required to for a pulse to propagate from the origin to position x is x / v.
So What happens at x at clock time t is what happened at the origin at clock time t - x / v.
At clock time t - x / v the point at the origin was at y position y = A sin (omega (t - x / v)).
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&#Good work. Let me know if you have questions. &#