Friday

March 6, 2015

March 6, 2015

Posted by **COFFEE** on Saturday, April 21, 2007 at 11:33pm.

(a) the listener is standing still and the tuning forks both move to the right at 30 m/s

(b) the tuning forks are stationary and the listener moves to the right at 30 m/s

I tried using:

f' = f [(v +/- vD)]/[(v +/- vS)]

f' = (329.6)[(343 - 0)/(343-30)]

f' = 31.6 Hz

Is this correct?

The results will be slghtly different for the two cases. You have the correct formula but did not use it properly.

For case (b), the two frequencies received are

329.6(343/313) and 329.6(343/373) = 361.2 and 303.1 and the beat frequency is 58.1 Hz.

For case (a), the frequencies are

329.6(313/343) and 329.6(373/343) = 300.8 and 358.4m and the beat frequency is 57.6 Hz.

ok thanks a lot!

**Answer this Question**

**Related Questions**

Physics - I did a lab in which we were to find the 1st resonance and 2nd ...

Physics - Two identical tuning forks have frequencies of 512 Hz. If one is held ...

Physics - Question about doppler effect: How do you solve when it gives you ...

Math/Physics - Two identical tuning forks have frequencies of 512 Hz. If one is ...

Physics (Waves) - You have 27 tuning forks that oscillate at close but different...

physics - What is the expected period of beats formed when two tuning forks ...

Physics - A tuning fork produces a sound with a frequency of 277 Hz and a ...

science - two tuning forks are struck at the same time. one tuning forkhas a ...

science - two tuning forks are struck at the same time. one tuning forkhas a ...

calculus - Interference Two identical tuning forks are struck, one a fraction of...