Doppler Effect; physics help please?
First find the velocity to produce waves of F = 485 behind the fork: λ = (u+v)/Fs → u/485 = (u+v)/512 → v = u(512/485 - 1) = 18.93 m/s The distance fallen when this frequency is first produced is h = v²/(2g) = 18.93²/(2*9.8) = 18.283 m. The time for this frequency to reach the release point is 18.283/340 = .05377 sec during which time the fork falls an additional height h' = v*t + ½gt² = 18.93*.05377 + ½*9.8*(.05377)² = 1.032 m Final answer: D = h + h' = 19.315 m
Physics - Doppler effect?
I'll assume the question should read "you are driving toward a cliff..." a. Yes, you hear a Doppler shift in the echo; specifically, you hear twice the shift that a person standing at the base of the cliff hears. This is because not only are you "compressing" the sound waves as you move forward, you are also racing through the reflection. (OK, maybe "racing" is too strong a word.) The first two sites give a pretty good treatment of Doppler. b. Although you are the source, it is like "moving target". This is so because you are moving toward your echo (exactly the principle bats use for echolocation, where they are the moving source). It might be easier to imagine that you are stationary and the cliff is moving toward you. And it might not. Check out the third cite, my favorite for the "moving target" example. c. The alien spacecraft (or the moving cliff!) merely exaggerates the effect.
Physics, Speed of Light and Doppler Effect?
A lidar (laser radar) gun is an alternative to the standard radar gun that uses the Doppler effect to catch speeders. A lidar gun uses an infrared laser and emits a precisely times series of pulses of infrared electromagnetic waves. The time for each pulse to travel to the speeding vehicle and return to the gun is measured. In one situation a lidar gun in a stationary police car observes a difference of 1.15 10-7 s in round-trip times for two pulses that are emitted 0.540 s apart. Assuming that the speeding vehicle is approaching the police car essentially head-on, determine the speed of the vehicle.
Physics, 9, hertz, doppler effect frequency?
vs = source emitting note = +30 (+ve taken along advaning source) F = 262 A) vo = observer on other train = - [18] F_ = apparent freq heard by observer v = speed of sound = advaning towards observer = reaching it = + 343 m/s F_ = F [v-vo]/[v-vs] ------------------------------- in this formula, everyone, v, vo, vs must move in the same ditection chosen as +ve. if one is in opposite direction, then resolve it along using -ve sign ------------------------- = 262[343 +18]/[343-30] F_ = 302.18 Hz ----------------- B) vo = +18 >> rest same F_ = 262[343 -18]/[343-30] = 272 Hz ========================= correction>.. i dis not read the meticulous wording case when first case observer has crossed the train vo = - 18 m/s v = - 343 m/s >. sound is reaching observer in opposite direction > so resolved F_ = 262[- 343 +18]/[- 343-30] F_ = 262[343 -18]/[343+30] = 228.28 Hz
Physics homework help? the doppler effect??
Alright, start with this Train 2 is the source of sound having a frequency of 500 Hz. and if it's moving towards train 1 that means it's postive so + 65 km/h . Now the observer on train one is also moving towards train 2 so + 140 km/h. Now you have the find what speed the sound travels through air during 1 km/h so v= 345 m/s / 1000m ( to get it to km) * 3600 sec ( to have 1 hour) = 1240 km/h now finally you can use your equation up there making sure to put 1240 km/h in v rather than 345m/s. Good luck Hope this helps
Doppler effect Problem?
I’ll take the speed of sound as c = 334m/s. Change the calculation if you are supposed to use a different value. For 100rpm, period T = 60s/100 = 0.6s v = ωr = (2π/T)r = (2π/0.6)*1 = 10.47m/s ___________ Highest frequency is when source moves towards student at 10.47m/s. . For a stationary observer with source moving towards them at speed v: Observed frequency f’ = f(c/(c-v)) = 400*(334/(334-10.47)) = 412.9Hz So the difference is 412.9 – 400 = 12.9Hz = 13Hz to 2 sig. figs.) ___________ Lowest frequency is when source moves away from student at 10.47m/s. For a stationary observer with source moving away from them at speed v: Observed frequency f’ = f(c/(c+v)) = 400*(334/(334+10.47)) = 387.9Hz So the difference is 400 – 387.9 = 12.1Hz (= 12Hz to 2 sig. figs.) ___________ Remember to change the calculation if your speed of sound is different.