- #1
Myr73
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A simple harmonic oscillator has total energy
E= ½ K A^2
Where A is the amplitude of oscillation.
E= KE+PE
a) Determine the kinetic and potential energies when the displacement is one half the amplitude.
b) For what value of the displacement does the kinetic energy equal the potential energy?
For a) I got ,
KE=?= 0.5mv^2
PE=?= mgy= 0.5 Kx^2
X= 0.5 A
0.5KA^2= 0.5mv^2 + 0.5Kx^2
0.5KA^2= 0.5mv^2 + 0.5K (0.5A)^2
KE= 0.5KA^2- 0.5K (0.5A)^2 = 0.5k(A^2- 0.5A^2)
PE= 0.5KA^2-0.5mv^2
And for b) all I have so far is
,Kinetic Energy= 1/2mv^2
Potential Energy = 1/2mx^2
So when KE=PE, then 1/2mv^2=1/2kx^2.
If you times by 2 then mv^2= kx^2 , therefore→ SquareRoot(mv^2/k)= x
But I don't know if that is correct or if it missing information
E= ½ K A^2
Where A is the amplitude of oscillation.
E= KE+PE
a) Determine the kinetic and potential energies when the displacement is one half the amplitude.
b) For what value of the displacement does the kinetic energy equal the potential energy?
For a) I got ,
KE=?= 0.5mv^2
PE=?= mgy= 0.5 Kx^2
X= 0.5 A
0.5KA^2= 0.5mv^2 + 0.5Kx^2
0.5KA^2= 0.5mv^2 + 0.5K (0.5A)^2
KE= 0.5KA^2- 0.5K (0.5A)^2 = 0.5k(A^2- 0.5A^2)
PE= 0.5KA^2-0.5mv^2
And for b) all I have so far is
,Kinetic Energy= 1/2mv^2
Potential Energy = 1/2mx^2
So when KE=PE, then 1/2mv^2=1/2kx^2.
If you times by 2 then mv^2= kx^2 , therefore→ SquareRoot(mv^2/k)= x
But I don't know if that is correct or if it missing information