course Phy 232
I plan on taking both test 1 and test 2 by the end of next week so I should be caught up by then.
Problem Number 1 An observer moves away from a stationary sound source with frequency 590 Hz at 26 m/s. If the speed of sound is 340 m/s, then what frequency will be heard by the obsever?
Solution: I tried to solve this using the formula 590=x(340/340-26) I then found x to 547.6 Hz.
This works. Be sure you know all the variations on the formulas for Doppler shift, and that you can also derive them (e.g., as in the intro problem sets).
Problem Number 2
Show that the function y(x,t) = .62 sin( 510 t - .52 x ) satisfies the wave equation, and give the velocity of the wave.
Solution: I know the wave equation is a second derivative equation, so I think I would just take the double derivative. My calculus is a little rusty so I kind of forget what exactly to do with the values inside the parenthesis.
You apply the chain rule, (f(g(x))) ' = g ' (x) * f ' (g(x)). For example, to get the x derivative:
( sin(510 t - .52 x) ) ' = (510 t - .52 x) ' * cos(510 t - .52 x) ) = -.52 cos(510 t - .52 x), where the ' refers to the x derivative.
We would then write
y_x = .62 ( -.52 cos(510 t - .52 x) ) = -.32 cos(510 t - .52 x).
The second y derivative would be found similarly; we would get
y_xx = -.32 ( -.52 ( -sin(510 t - .52 x) ) ), which would be simplified to -.16 sin(...etc.) .
You should review the rules for taking derivatives, which is something you will likely have to do on the test.
Problem Number 3
What is the fundamental frequency of a longitudinal standing wave in an aluminum rod of length 8 meters, balanced at its midpoint, if a longitudinal disturbance travels at 5000 m/s?
Solution: I know the unit of frequency is Hz which is inverse seconds. So I think I should first find the seconds by taking 5000/8 = 625 I then think I would take 1/625 = 0.0016 Hz. I’m not sure if I did this correctly I didn’t really look up the answer I just went with basic knowledge. I’m also not sure if the aluminum part has anything to do with solving this problem. I don’t believe is does though.
Use units.
5000 m/s / (8 m) = 625 s^-1 or 625 Hz. This is the frequency.
1 / (625 Hz) = .016 sec, which is the period of the wave.
Work through the Introductory Problem Sets.
Problem Number 4
A person 1.35 meters high in front of a converging lens whose focal length is 1.87 meters. A real inverted image of forms 5 m on the other side of the lens.
• How far is the person from the lens?
• How large is the image?
• Sketch a diagram explaining how the image forms.
Solution: I think to solve the first question you would take (1.87)*(1.35)/5 = 0.505 meters. I’m not real sure how to calculate how large the image is I looked for the equation I even tried google, but it was a little confusing. I have drawn a diagram, but I am not sure if it was what you were expecting.
1/f = 1/i + 1/ o. You apply this by identifying the focal length f, the image distance i and the object distance o. In this case you are given the image and object distances.
THe magnification (not requested on this problem but frequently requested on tests) is related to the ratio of image and object distances in a simple and obvious way that is explained in the text, with diagrams to help make sense of it.
Problem Number 5
How many times louder is a sound which measures 79 decibels and a sound which measures 27 decibels?
Solution: I’m sure this is more difficult than it seems, but by just dividing 79/27 = 2.93 times louder. I doubt that’s right, but it’s possible.
You have to know the definition of the decibel. A decibel scale is a logarithmic scale; dB = 10 log ( I / I0 ).
Problem Number 6
At clock time t the y position of the left-hand end of a long string is y = 1.19 cm * sin ( ( 5 `pi rad/s) t ). The string extends along the x axis, with the origin at the left end of the string, which is held under tension 51 Newtons and has mass per unit length 16 grams / meter. The motion of the left end creates a traveling wave in the string. How much energy is there per unit length of this wave? What is the precise nature of this energy?
Solution: I don’t really understand the question. I think it just asks for you to take 51/16 = approx. 3.19 units per length of the wave. This doesn’t seem right, but I’m not sure. The precise nature of the energy? Does this mean what is the energy doing exactly?
This comes from the introductory problem sets. The energy in the traveling wave is equally distributed between kinetic and potential energies, in a way explained in the solution to the corresponding problem.
Problem Number 7
A string of length 9 meters is fixed at both ends. It oscillates in its third harmonic with a frequency of 198 Hz and amplitude .83 cm. What is the equation of motion of the point on the string which lies at 3.3 meters from the left end? What is the maximum velocity of this point?
Solution: What exactly does it mean by “in its third harmonic?” I found a problem similar to this and I believe you find the answer the same way. The equation of motion would be yMax=.83cm*sin(3488/m*(x)) where x would be 3.3 meters. So I guess to find the maximum velocity you would just plug in the 3.3 which would give you 1.68 *10^-4. I don’t think this is right there must be an equation or something.
The harmonics of a string are explained in detail in your text and in the Introductory Problem Sets.
The first three harmonics of a string fixed at both ends have wavelengths 2 L, 2 L / 2 and 2 L / 3.
From the frequency and wavelength you can figure out the propagation velocity.
At the 3.3 m point the amplitude would be y_max = .83 cm sin( k x ), where k is the wave number and x is your 3.3. meters. k is obtained from the wavelength.
3488 is the angular frequency of the oscillation, which is not relevant to calculating the amplitude of the motion at x = 3.3 m.
However having calculated this, the equation of motion of the point is y(t) = y_max sin(omega * t), where omega is your angular frequency omega = 3488 rad / sec.
Taking the derivative of the y(t) function you get the velocity function for the transverse oscillation at this point.
Problem Number 8
A string defines the x axis, with the origin at the left end of the string. The string has tension 19 Newtons and mass per unit length is 14 grams / meter. The left-hand end of the string is displaced perpendicular to the string in a cyclical manner in order to create a traveling wave in the string. At clock time t the position of the left-hand end of a long string is y = .54 cm * sin ( ( 9 `pi rad/s) t ).
• What equation describes the displacement a point 16.7 meters down the string as a function of clock time?
• If the position of the left-hand side is x = 0, what is the equation for the displacement as a function of clock time at arbitrary position x?
• University Physics: Show that your equation satisfies the wave equation.
• What is the equation for the shape of the string at clock time t = .042 sec?
Solution: I ‘m not sure if the 1st question wants an equation or a answer. I think you would plug in the 16.7 meters into the given equation. The second equation would just be y(t) = 0 I believe. I also don’t really see what is meant by arbitrary position x. To find the 4th answer I think this would just get plugged into the equation. If all these questions do just want formulas and not answers then I don’t really understand why any information was given in the explanation.
The first question wants the equation of motion for a specific point on the wave. The equation will be much like the equation of motion given for the left end of the string, but with a time delay (it takes a certain specific time for the disturbance to move 16.7 meters down the string; you will need to calculate the speed of the pulse and use it to find the delay).
The first question asks about a specific point on the string. The second asks you to generalize your analysis to an arbitrary point x. This gives you the general equation for y(x, t), the function that describes the transverse displacement of a point on the wave at a given position and a given instant in time. Having obtained this equation, you take its second x derivative and its second y derivative, and relate them using the wave equation. Your result should be consistent with the propagation velocity you calculated earlier.
If you plug a specific value of t into your equation for y(x, t), then t is no longer variable and you have a function y(x) for that specific t. The graph of this function describes the shape of the wave at this instant.
Except for the part about satisfying the wave equation, this is an Introductory Problem Set problem.
.Problem Number 9
If a traveling wave has wavelength 2.3 meters and the period of a cycle of the wave is .035 second, what is the propagation velocity of the wave? What is its frequency
Solution: I’m not real sure what propagation velocity is, but the frequency is the inverse of time. The frequency would then be 1/.035 = 28.6 Hz.
This is an early Introductory Problem Set problem. Your text also gives you the means to answer this question.
See my notes, and be sure you master the concepts and procedures of the Introductory Problem Sets. Sometimes is looks like you are trying to find formulas for situations that are easier to simply reason out based on the concepts, without which it would in any case be difficult to correctly apply the formulas.
You probably need to spend a couple of hours reviewing the basic rules for derivatives of composite functions, especially composites involving trigonometric and exponential functions, both of which can occur on this test.
I'll be glad to answer additional questions.
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