#$&* course Phy 122 6:20 pm June 28 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.
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Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Intensity is also involved in this process ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: use the formula from earlier, dB = 10 log (I / I_threshold) and solve for I log (I / I_threshold) = dB / 10 I / I_threshold = 10 ^(120 / 10) = 12 I = I_threshold * 10 ^12 I = 10 ^ -12 watts / m ^2 * 10 ^12 = 1 watt / m ^2 (dB = 20 watts) I = I_threshold * 10 ^(20 / 10) = 10 ^-12 watts / m ^2 * 10 ^2 = 10 ^-10 watts / m ^2 120 dB is 10 billion times as intense confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I. We get log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12and I = I_threshold * 10^12. Since I_threshold = 10^-12 watts / m^2, we have for dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense. A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]} = 10 { log(I_1) - log(I_2)} = 10 log(I_1 / I_2). So we have 120 - 20 = 100 = 10 log(I_1 / I_2) and log(I_1 / I_2) = 100 / 10 = 10 so that I_1 / I_2 = 10^10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Openstax: Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0ºC . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: wavelength in water will be about 5 times that in air because the speed of sound in water is about 4 times that in air and all of the peaks will be distrubuted over about 5 times in the water confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using propagation speeds 340 m/s and 1600 m/s for sound in water and air, respectively: The frequency of a particular sound is determined by the frequency at which the vocal cords vibrate, and is the same in water as in air. Since the speed of sound in water is roughly 5 times that in air, the same number of peaks will be distributed over about 5 times the distance in water as opposed to in air. So the wavelength in water will be about 5 times that in air. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: Openstax: The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: log ( I / I_0) = dB / 10. (dB = 91) log (I / I_0) = 91/10 = 9.1. I = 1.3 * 10 ^9 * I_0 = 1.3 * 10 ^9 * (10^-12 watts / m^2) = 1.3 * 10 ^-3 watts / m ^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: dB = 10 log( I / I_0), so log( I / I_0) = dB / 10. In this case dB = 91 so log(I / I_0) = 91/10 = 9.1. It follows that I / I_0 = 10^9.1 = 1.3 * 10^9, approx.. so that I = 1.3 * 10^9 * I_0 = 1.3 * 10^9 * (10^-12 watts / m^2) = 1.3 * 10^-3 watts / m^2, or about .0013 watts / m^2. Your Self-Critique: ok Your Self-Critique Rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!