A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what

Question

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what displacement from the equilibrium position is the particle's speed equal to one half its maximum speed. What fraction of its maximum speed does the particle have when it is halfway to its amplitude?
The actual answer is 2.60 cm and v = 0.866 vmax

First off, I keep coming up with 2.1 cm following the other answers here. I thought I would use the
1/2 kA^2 = 1/2 mv^2 + 1/2 kx^2 but we weren't give the spring constant. The other formulas used a version of 1/2 kA^2 = 12 kx^2 (but that didn't make sense or give the right answer). The book uses 1/2 kA^2 (but has the spring constant) to find energy then solves for velocity. I'm close, but no prise. Thanks for the help!

Explanation / Answer

Above answer is correct. use following steps:

V(max)= ((k/m)^.5)*A ,

1/2 kV^2 = 1/2 mv^2 + 1/2 kx^2 ,

put value of V(max) and take v= V/2 ,

you will ended with

3/4 kV^2 = 1/2 kx^2 ,

x= .867*A = 2.6 cm,

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