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January 31, 2015

January 31, 2015

Posted by **Trace** on Thursday, March 3, 2011 at 10:42pm.

of her. At the bottom of her swing when she is moving toward the whistle, she hears the higher

pitch, and at the bottom of her swing, when she is moving away from the swing she hears a lower

pitch. The higher pitch has a frequency that is 5.0% higher than the lower pitch. What is the

speed of the child at the bottom of the swing?

I think I need to use the equation

freq observed=[(1-v child)/v sound]freq sound

I know the v sound =343 m/s but I have no idea even where to start on this question.

- physics -
**Trace**, Thursday, March 3, 2011 at 11:21pmokay, so I set it up as a ratio:

[(1-v child)/v]*freq sound

OVER

[(1+v child)/v]*freq sound

ALL THAT =1.05

So I have 1.05=[(1-v child)/343]/[(1+v child)/343]

But I can't figure out how to manipulate this equation. I keep getting the v child as 341 m/s, and I know that kid is not moving that fast! How do I finish this equation?

- Never mind I got it!! -
**Trace**, Thursday, March 3, 2011 at 11:28pmI guess I gave up too soon!

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