Post a New Question


posted by on .

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 - ,

    okay, so I set it up as a ratio:
    [(1-v child)/v]*freq sound
    [(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!! - ,

    I guess I gave up too soon!

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question