Recent content by pot

That makes sense. Thank you so much for the help!

Oh, I see! So for 2) I would have to use the formula twice again so it would be f(mos. hears) = (10,300)[(340 - 1)/340] = 10,270 Hz and then: f(bat still) = (10,270)[340 / (340+1)] = 10,240 Hz

So for 1) I would have to use the formula twice? Like this: f(mos. hears) = (10,300)[(340 - 1)/(340 - 1.05)] = 10,302 Hz Then: f(bat hears) = (10,301)[(340 + 1.05)/(340 + 1)] = 10,303 Hz 2) f(bat hears) = (10,301)[340/(340 + 1)] = 10,271 Hz

For 3) Yes, as the bat can detect a change in frequency as low as 1.00 Hz, and the frequency in both 1) and 2) change by more than 1 Hz (from 10,300 Hz to less than 10,299 Hz). And about the Doppler shift formula, I'm confused. Can you explain how I'm not using it right?

Oops, sorry about that. I still included the frequency value in my calculations. I fixed it now. I thought the Doppler shift equation would be used when calculating frequencies that are effected by other speeds. Do you mean I have chosen the wrong +/-? Rereading the question now, I think 1)...

Hi, Would 1) be f' = (10,300)[(340 - 1.00) / (340 - 1.05)] = 10,302 Hz 2) f' = (10,300)[(340 - 1.00) / (340)] = 10,270 Hz 3) Yes? As the frequency changes more and less than 1.00 Hz. I am unsure and would appreciate any help. Thanks!