Homework Help: physics

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

A child swinging on a swing set hears the sound of a whistle that is being blown directly in front
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 11:21pm
okay, 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 11:28pm
I guess I gave up too soon!

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