At what displacement from the equilibrium is the total energy of a simple harmonic oscillator one - sixths KE and five - sixth PE? Answer should be in terms of the amplitude A.
Let A be the full amplitude displacement and k be the sping constant.
The maximum total energy is
Et = (1/2) k A^2
The potential energy for other displacements is
Ep = (1/2) k x^2
Ep/Et = 5/6 when (x/A)^2 = 5/6
x/A = 0.91287
To find the displacement from the equilibrium at which the total energy of a simple harmonic oscillator is one-sixth KE (kinetic energy) and five-sixth PE (potential energy), we need to understand the relationship between the total energy, kinetic energy, and potential energy of a simple harmonic oscillator and its displacement.
In a simple harmonic oscillator, the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE) at any given displacement x from the equilibrium position.
E = KE + PE
The kinetic energy of a simple harmonic oscillator is given by:
KE = (1/2) m v^2
where m is the mass of the oscillator and v is its velocity.
The potential energy of a simple harmonic oscillator is given by:
PE = (1/2) k x^2
where k is the spring constant and x is the displacement from the equilibrium position.
According to the problem statement, the total energy is one-sixth KE and five-sixth PE. Let's assume the total energy is E_0. Therefore, we have:
E_0 = (1/6) KE + (5/6) PE
Now, we can substitute the expressions for KE and PE into the equation:
E_0 = (1/6) (1/2) m v^2 + (5/6) (1/2) k x^2
However, we need to relate the velocity (v) with the displacement (x). In a simple harmonic oscillator, the velocity is related to the displacement through the angular frequency (ω) and amplitude (A) by the equation:
v = ω A
The angular frequency (ω) is related to the mass (m) and spring constant (k) by the equation:
ω = sqrt(k / m)
Substituting this relation into the expression for velocity (v) gives:
v = sqrt(k / m) A
Now we can substitute this expression for v and simplify the equation:
E_0 = (1/6) (1/2) m (sqrt(k / m) A)^2 + (5/6) (1/2) k x^2
E_0 = (1/6) (1/2) m (k / m) A^2 + (5/6) (1/2) k x^2
E_0 = (1/6) (1/2) k A^2 + (5/6) (1/2) k x^2
E_0 = (k / 12) A^2 + (5/12) k x^2
Now, we can set this equation equal to the total energy (E_0) and solve for the displacement (x):
E_0 = (k / 12) A^2 + (5/12) k x^2
Solving for x:
(k / 12) A^2 + (5/12) k x^2 = E_0
(5/12) k x^2 = E_0 - (k / 12) A^2
k x^2 = 12(E_0 - (k / 12) A^2) / 5
x^2 = 12(E_0 - (k / 12) A^2) / (5k)
Taking the square root of both sides:
x = sqrt(12(E_0 - (k / 12) A^2) / (5k))
So, the displacement from the equilibrium position at which the total energy is one-sixth KE and five-sixth PE is given by x = sqrt(12(E_0 - (k / 12) A^2) / (5k)).
Please note that without specific values for the total energy (E_0), mass (m), spring constant (k), and amplitude (A), we can't provide a numerical answer.