#$&* course Phy 122 015. `Query 13
.............................................
Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand the concept now. I didn’t offer enough proof. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I=Ithreshold*10^12=1watt dB for 20 dB=10^-10 watts 1/10^-10=10^10 watts/m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!