If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

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.

 

013. `Query 11

 

Question: `query introductory set six, problems 15-18

 

How do we determine the energy of a traveling wave given the amplitude and frequency of the wave, and the mass of the segment of string in question?

 

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Confidence Rating:

 

Given Solution:

`aSTUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2

 

INSTRUCTOR RESPONSE:

 

** You should understand the way we obtain this formula.

 

We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results.

 

Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. **

 

STUDENT QUESTION

 

I found the equation above in the book and understand, but my answer was based on problem 17.


I don’t understand why the solution is different?

INSTRUCTOR RESPONSE

 

Your solution was in terms of omega; the equation you quote was an intermediate step in the solution process.  That equation is accurate, but it doesn't express the result in terms of the quantities given here. The given information didn't include omega, but rather gave the frequency of the oscillation.

 

So putting everything in terms of f, A and mass m:


omega = 2 pi f, so vMax = omega * A = 2 pi f * A and

 

total energy = .5 m vMax^2 = .5 m * ( 2 pi f * A)^2,

 

which when expanded is equal to the expression in the given solution.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

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Question: `qIf the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase?

 

 

 

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Given Solution:

`a** the question here is whether the far ends of the strings are at the same phase of motion, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths.

 

Alternatively, the pulse in the longer string will be 'behind' the pulse in the shorter by the time required to travel the extra length. If we know the frequency we can determine whether this 'time difference' corresponds to a whole number of periods; if so the ends will oscillate in phase **

STUDENT QUESTION:

 

The pulse of the longer string will take obviously longer than the shorter string but if the
frequency is the same they will be oscillating at the same rate. Im not sure if I truly understand.

INSTRUCTOR RESPONSE:

 

If the strings are of the same length then, given the specified conditions, their ends will oscillate in phase. When a peak arrives at the end of one string, a peak will arrive simultaneously at the end of the other.

If you trim a little bit off the end of one of the strings, this won't be the case. When a peak arrives at the end of the untrimmed string, the peak will have already passed the end of the trimmed string, which is therefore oscillating ahead of the phase of the end of the untrimmed string.

The ends of both strings will therefore be oscillating with the same frequency, but out of phase.

 

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Question:    Openstax:  How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?

 

 

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Given Solution:  Crests of the wave travel at 5 m/s and are separated by one wavelength, or 40 m.  Thus a crest will pass the boat every 40 m / s (5 s) = 8 seconds.

 

Your Self-Critique:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Your Self-Critique Rating:



 

Question:    Openstax:  A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce?

 

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Given Solution:  A wave with frequency 199 Hz will come in and out of phase with a wave having frequency 203 Hz four times every second.  So the beat frequency is 4 Hz.

 

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Question:    Openstax:  (a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, they compare the arrival times of S- and P-waves, which travel at different speeds. Figure 16.48) If S- and P-waves travel at 4.00 and 7.20 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.)

 

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Confidence Rating:

 

Given Solution: 

 

One wave travels 3.20 km/s faster than the other.  If the distance to the epicenter is `ds, then the time between the arrival of the two waves will be `dt = `ds / (3.20 km/s).

 

If `dt is uncertain by 0.10 second, then `ds will be uncertain by 3.20 km/s * 0.10 s = .32 km, or 320 meters.

 

Your Self-Critique:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Question: `qGeneral College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges?

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Given Solution:

`a At 3 * 10^8 m/s:

 

a frequency of 550 kHz = 550 * 10^3 Hz = 5.5 * 10^5 Hz will correspond to a wavelength of 3 * 10^8 m/s / (5.5 * 10^5 cycles / sec) = 545 meters.

 

a frequency of 1600 kHz = 1.6* 10^6 Hz will correspond to a wavelength of 3 * 10^8 m/s / (1.6 * 10^6 cycles / sec) =187 meters.

 

The wavelengths for the FM range are calculated similarly.

 

a frequency of 88.0 mHz= 88.0 * 10^6 Hz = 8.80 * 10^7 Hz will correspond to a wavelength of 3 * 10^8 m/s / (8.80 * 10^7 cycles / sec) = 3.41 meters.

 

The 108 mHz frequency is calculated similarly and corresponds to a wavelength of 2.78 meters.

 

STUDENT QUESTION

 

I don’t understand where the v came from? How did you get the velocity to work the problem?

INSTRUCTOR RESPONSE

 

This is electromagnetic radiation. Its propagation velocity is the speed of light.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

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Question: `qGeneral College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz?

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Given Solution:

`aThe fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are

 

fundamental frequency = 440 Hz

First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz

Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz

Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz

STUDENT QUESTION

 

Where does the 4th come from? Didn’t see that in the notes?

INSTRUCTOR RESPONSE

 

The pattern of the frequencies should be clear from the way the wavelengths are determined. The pattern follows the obvious pattern of the first three harmonics.

The nth harmonic will contain n half-wavelengths in the distance covered by a single half-wavelength of the fundamental. So its frequency will be n times higher than the fundamental. The frequency of the nth harmonic is therefore n times that of the fundamental.

 

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Question: `qGeneral College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source?

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Confidence Rating:

 

Given Solution:

 

The wave is assumed spherical so its surface area increases as the square of its distance.

 

Its intensity, which is power / surface area, therefore decreases as the square of the distance.

 

So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great.

 

2300 times the original intensity is

In symbols, we express the fact that the intensity is inversely proportional to the area by indicating that the ratio of intensities is the reciprocal of the ratio of areas:

Since 1 / (A2 / A1) = A1 / A2 we can express the same thing as follows:

Now the ratio of the areas of two sphere is equal to the square of the ratio of their radii:  A1 / A2 = (r1 / r2)^2.  It follows that

In the present case, then, we easily obtain

Since r1 = 48 km and r2 = 1 km we get

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

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Question: `qAt what rate did energy pass through a 5.0 m^2 area at the 1 km distance?

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Confidence Rating:

 

Given Solution:

`aThrough a 5 m^2 area the rate of energy passage is therefore 4.6 * 10^9 J / (m^2 s) * 5.0 m^2 = 2.3 * 10^10 J / s, or 23 billion watts.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

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