query assignment 22

course phy202

~PS˫ўassignment #022

022. `Query 11

Physics II

04-02-2008

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10:30:36

Query introductory set six, problems 15-18

how do we determine the energy of a standing wave given the amplitude and frequency of the wave and the mass of the string?

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RESPONSE -->

The formulas of SHM apply. Each particle will have energy equivalent to the max KE which is 1/2*m*maxV^2

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10:31:46

STUDENT 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. **

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RESPONSE -->

ok

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10:38:37

If 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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RESPONSE -->

If the wave velocity of the two strings are identical and they are not fixed at the other end, then the waves should remain in phase with each other.

If the other ends are not fixed, then we would haveto detemine how the difference in length relaets to the wavelength and whether the difference in length is equal to some multiple of the wavelength so that the ends of both strings end at the same phase of the wave form and thus move in unison.

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10:39:08

** 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 **

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RESPONSE -->

ok

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10:57:17

General 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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RESPONSE -->

3.010^8 m/s / 550 10^3 cycles/sec = lambda of 545 m

3.0 10^8 m/s / 1600 10^3 cycles/sec = lambda of 187 m

3.0 10^8 m/s / 88 10^6 cycles/sec = lambda of 3.4 m

3.0 10^8 m/s / 108 10^6 cycles/sec = lambda of 2.8 m

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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10:57:44

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.

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RESPONSE -->

ok

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:03:17

General 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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RESPONSE -->

lambda 1 = 440 hz

lambda 2 = 880 hz

lambda 3 = 1320 hz

etc ......

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:03:34

The 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

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RESPONSE -->

ok

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:15:18

General 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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RESPONSE -->

for a spherical wave, the power decreases with the inverse square of the radius so:

2 10^6 J * 48km^2 = strength at source = 4.6 10^15J / 1km^2 = energy at 1 km from source = 4.6 10^11 J

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:16:29

The wave is assumed spherical so its surface area increases as the square of its distance and its intensity, which is power / surface area, decreases as the square of the distance. So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great, or 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s).

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RESPONSE -->

ok

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:19:14

At what rate did energy pass through a 5.0 m^2 area at the 1 km distance?

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RESPONSE -->

It would be 5 times the energy per m^2 or 4.6 10^9 J/m^2 * 5m^2 = 23 10^9 J

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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11:20:00

Through 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.

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RESPONSE -->

Made a math error but I understand

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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