A simple harmonic oscillator of amplitude A has a total energy E . (a) Determine

Question

A simple harmonic oscillator of amplitude A has a total energy E.

(a) Determine the kinetic energy when the position is one-third the amplitude. (Use any variable or symbol stated above as necessary.)
K =

(b) Determine the potential energy when the position is one-third the amplitude. (Use any variable or symbol stated above as necessary.)
U =

(c) For what values of the position does the kinetic energy equal one-half the potential energy? (Use any variable or symbol stated above as necessary.)
x = ±

(d) Are there any values of the position where the kinetic energy is greater than the maximum potential energy?

YesNo    

Explain.

Explanation / Answer

a)

When it is at one third of amplitude, the PE stored in the spring , PE = 0.5*k*(A/3)^2 = k*A^2/18  

Total energy = 0.5*k*A^2 = kA^2/2

So, at that point, its PE energy = Total energy/ 9

So, KE = total energy - PE = E - E/9 = 8*E/9 <------- answer

b)

U = E/9 <---- as explained above

c)

KE = E/2

So, PE = E - E/2 = E/2

So, 0.5*k*x^2 = (0.5*k*A^2)/2

So, x = +/- (A / sqrt(2)) <---------- A = amplitude

d)

Maximum PE = E

Now, the maximum value either PE or KE an take = E

So, answer is NO

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