course PHY 202 ?????????????vassignment #024
......!!!!!!!!...................................
05:34:10 query experiment to be viewed and read but not performed: transverse and longitudinal waves in aluminum rod what is the evidence that the higher-pitched waves are longitudinal while the lower-pitched waves are transverse?
......!!!!!!!!...................................
RESPONSE --> I could tell that the higher pitched waves were longtitudinal because when you put your finger on the end of the aluminum rod, the higher pitched waves were not damped. This could mean that your finger was not interfering with the frequency, thus the waves were moving along the direction of the wave's motion. I know that there were transverse waves presend because of the lower pitches, but I am not complelety sure how I could prove that they were transverse
.................................................
......!!!!!!!!...................................
05:37:47 STUDENT 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. **
......!!!!!!!!...................................
RESPONSE --> ok, we would need to measure velocity to be more accurate but we can tell that since the higher pitched waves were damped more easily by touching the end of the rod than the lower pitched ones, than we can conclude that the lower pitched waves were transverse and the finger was not interfering with the frequency, because they were perpendicular to the motion of the wave itself.
.................................................
......!!!!!!!!...................................
05:58:34 Query 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.
......!!!!!!!!...................................
RESPONSE --> Using the formula: 10log(I/Isubscript0) Thus I = 120 and I subscript 0= 1.0E-12. Then 10log(120/1.0E-12)=1.2E15 I=20, I subscript 0=1.0E-12. Then 10 log(20/1.0E-12)=2E13
.................................................
......!!!!!!!!...................................
06:04:58 The 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.
......!!!!!!!!...................................
RESPONSE --> "" I subscript 0""should actually be written as I_threshold.
.................................................
......!!!!!!!!...................................
06:09:00 Query gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?
......!!!!!!!!...................................
RESPONSE --> Well we know that the speed of sound in air at room temp(21C) is 343 m/s. So given the formula of L=v/2f, we have L=343/2(262)=.65 meters
.................................................
......!!!!!!!!...................................
06:14:01 GOOD STUDENT SOLUTION First we must determine the velocity of the sound waves given the air temperature. We do this using this formula v = (331 + 0.60 * Temp.) m/s So v = (331 + 0.60 * 21) m/s v = 343.6 m/s The wavelength of the sound is wavelength = v / f = 343.6 m/s / (262 Hz) = 0.33 meters. So 262 Hz = 343.6 m/s / 4 * Length Length = 0.33 meters f = v / (wavelength) 262 Hz = [343 m/s] / (wavelength) wavelength = 1.3 m. So the wavelength is 1.3 m. If it's an open pipe then there are antinodes at the ends and the wavelength is 2 times the length, so length of the the pipe is about 1.3 m / 2 = .64 m, approx.. Had the pipe been closed at one end then there would be a node and one end and an antinode at the other and the wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.3 m / 4 = .32 m. **
......!!!!!!!!...................................
RESPONSE --> This is an open organ pipe, not a closed one, so I don't understand why the formula multiplying 4(length) was used. Other than that, I understand the process.
.................................................
~?????????q??w??assignment #024 024. `Query 13 Physics II 07-20-2007"