Question 16

[mvenus9292913]

Ok, maybe I'm just reading this explanation wrong (which is entirely possible... I haven't been thinking straight today)...

Therefore, the observed frequency is greater which means that the wavelength is lower from the observers point of view (from velocity = frequency x wavelength). Thus the wavelength from the source (= the original wavelength) is higher. As long as the environment (i.e. temperature) is relatively constant, the velocity of sound is constant (PHY 7.1.2, 8.2, 8.5).

So, the answer would be a smaller wavelength and the same velocity, right?

That would be B, not C.

[mcat_premed3832]

Not exactly. The reflected wave does indeed have a lower wavelength; however, the question is asking about the relative wavelength of the original wave.

[mvenus9292913]

Oh, I see it now. I said it was possible that I was just reading it wrong...

[asfino12217]

Why isn't it -Vo and Vs=0 since the source of the sound is going no where and the observer is coming closer?

[nedaa.asba6809]

Just to clarify that v = speed of the observer (o) or the source (s).

We are told that the observer is stationary (so the observer has no velocity, vo = 0).

We are told that an object (emitting a sound) is approaching with velocity v (so vs cannot be 0).

[admin]

Bump.