#$&* course PHY 232 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
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): ------------------------------------------------ Self-critique Rating:ok ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the problem requires as to find out wether the far ends of the strings have the same phase motion, we then have to put into account the wavelength, and since frequecy can give us wavelength, then we can use the wavelength to find out the phase motion. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Am not sure if my solution makes any sense am still alittle bit unsure as to how the wavelength will be used to determine the phase motion. ------------------------------------------------ Self-critique Rating:Ok
.............................................
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): ------------------------------------------------ Self-critique Rating:Ok ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we already know that wavelength is equal to double its length, we can fit the harmonics into the length of the string. For example since we are given the fundamental frequecy=440Hz then for the 2 hamornic frequecy = 2*440Hz=880Hz 3 hamornic frequecy = 3*440Hz=1320Hz 4 hamornic frequecy = 4*440Hz=1760Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are fundamental frequency = 440 Hz First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz STUDENT QUESTION Where does the 4th come from? Didn’t see that in the notes? INSTRUCTOR RESPONSE The pattern of the frequencies should be clear from the way the wavelengths are determined. The pattern follows the obvious pattern of the first three harmonics. The nth harmonic will contain n half-wavelengths in the distance covered by a single half- wavelength of the fundamental. So its frequency will be n times higher than the fundamental. The frequency of the nth harmonic is therefore n times that of the fundamental. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:Ok ********************************************* Question: `qGeneral College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we already know that intensity is inversely proportion to Area then have to find out how much is in 1km=(48km/1km)^2=2304 then we can use it to solve for the intensity on 1km as follows = 2304*(2*10^6 J/m^2s)=4.6*10^9 J/m^2s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 intensity at 1 km = 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s). 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: I2 / I1 = 1 / (A2 / A1) Since 1 / (A2 / A1) = A1 / A2 we can express the same thing as follows: I2 / I1 = A1 / A2 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 I2 / I1 = (r1 / r2)^2. In the present case, then, we easily obtain I2 = I1 * (r1 / r2)^2 Since r1 = 48 km and r2 = 1 km we get I2 = I1 * (48 km / (1 km))^2 = 2300 * I1, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qAt what rate did energy pass through a 5.0 m^2 area at the 1 km distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we are given the area we can say that the rate of energy will through it will be (4.6*10^9 J/m^2s)(5.0m^2)=2.3*10^10 J/s 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): ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!