As we know that KE and PE is same at a given position
so we will have as a function of position given as
[tex]KE = \frac{1}{2}m\omega^2(A^2 - x^2)[/tex]
also the PE is given as function of position as
[tex]PE = \frac{1}{2}m\omega^2x^2[/tex]
now it is given that
KE = PE
now we will have
[tex]\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2[/tex]
[tex]A^2 - x^2 = x^2[/tex]
[tex]2x^2 = A^2[/tex]
[tex]x = \frac{A}{\sqrt2}[/tex]
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
[tex]KE = \frac{1}{2}m\omega^2(A^2 - x^2)[/tex]
now to find the fraction of kinetic energy
[tex]f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}[/tex]
[tex]f = \frac{A^2 - (\frac{A}{3})^2}{A^2}[/tex]
[tex]f_k = \frac{8}{9}[/tex]
now since total energy is sum of KE and PE
so fraction of PE at the same position will be
[tex]f_{PE} = 1 - f_k[/tex]
[tex]f_{PE} = 1 - (8/9) = 1/9[/tex]
The fraction of kinetic energy and fraction of Kinetic energy and Potential energy is mathematically given as
a) [tex]x=\frac{A}{\sqrt{2}}[/tex]
b) F=8/9
c) Fp=1/9
Generally the equation for the Kinetic energy and Potential energy as is mathematically given as
KE = 1/2mw^2(A^2 - x^2)
and
PE = 1/2mw^2x^2
With
K.E and P.E equal we have[tex]x=\frac{A}{\sqrt{2}}[/tex]
b) for when x = A/3
Generally the equation for the Kinetic energy is mathematically given as
KE = 1/2mw^2(A^2 - x^2)
herefore, the fraction of Kinetic energy is
[tex]f = \frac{KE}{TE}[/tex]
[tex]f=\frac{A^2 - x^2}{A^2}[/tex]
f = 8/9
Therefore, fraction of PE is
Fp=1-f
Fp=1-(8/9)
Fp=1/9
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