Arailroad train is traveling at 32 m/s in still air. the frequency of the train whistle is 350 hz. what is the wavelength of the sound and the frequency of the sound heard by the listener in front of the locomotive?

40. 383 Hz, 0.9 m

6. 20.78 m/s

40. You know the pitch will be increased when the train is moving toward you. The only answer choice that is greater than 350 Hz is the one shown above.

Using the formula shown in the attachment, we compute the observed frequency to be ...

... f' = (343 m/s)/(343 m/s -32 m/s)×(350 Hz) ≈ 1.102893×350 Hz ≈ 386 Hz

The wavelength is given by ...

... λ = v/f = (343 m/s)/(386 Hz) ≈ 0.8886 m ≈ 0.9 m

_____

Comment on the answer choice

If you erroneously use the frequency modification formula f'=(v+vo)/v·f, then you get 382.7 Hz ≈ 383 Hz. We suspect an error in the answer key. Have your teacher show you how to work this problem.

__________

6. The velocity can be computed from the acceleration and distance covered by ...

... v = √(2ad)

... v = √(2·9.8 m/s²·22 m) = (√431.2) m/s ≈ 20.77 m/s

The closest answer among those offered is 20.78 m/s. Again, you should have your teacher show you how to work this problem to arrive at the offered answer.

One

I putting this up because I have answers. This is another doppler effect problem.

Givens

Formula

Solution

v= f * wavelength

v = 343 m/s

f = 386 Hz

343 = 386 * wavelength

343 / 386 = wavelength

wavelength = 0.881 m

386, 0.9 C

Problem Two

Givens

Formula

Solution

A

Note to get 20.78 they used 9.8 but then there should only be 2 places and the answer (rounded) is 21. Otherwise the answer is 20.77

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//