In a spring-mass system, the total mechanical energy is the sum of the kinetic energy and potential energy. At any point in the oscillation, when the kinetic energy and potential energy are equal, the total mechanical energy is constant. In this scenario, we can set up an equation to find the distance from the equilibrium position where this occurs.
Let's denote the equilibrium position as x = 0. The amplitude of the oscillation is given as 6√2. The total mechanical energy (E) can be calculated using the formula:
E = (1/2)kA^2
Where k is the spring constant and A is the amplitude of oscillation. Since the amplitude is given as 6√2, we have:
E = (1/2)k(6√2)^2 E = 72k
Now, let's consider a general point in the oscillation where the distance from the equilibrium position is x. At this point, the potential energy (PE) is given by:
PE = (1/2)kx^2
And the kinetic energy (KE) is given by:
KE = (1/2)mv^2
In this system, the maximum potential energy occurs at the amplitude (A), and the maximum kinetic energy occurs at the equilibrium position (x = 0). At the point where both kinetic energy and potential energy are equal, we have:
(1/2)kx^2 = (1/2)mv^2
Since the mass (m) and spring constant (k) are constants, we can equate the potential energy and kinetic energy expressions:
kx^2 = mv^2
Now, let's substitute v with its expression in terms of x using the conservation of mechanical energy:
(1/2)kA^2 = (1/2)mv^2
We already calculated the total mechanical energy (E) as 72k:
72k = (1/2)mv^2
Now, let's substitute v with its expression in terms of x:
72k = (1/2)m(dx/dt)^2
Since dx/dt represents the velocity (v), we can further simplify:
72k = (1/2)m(d/dt)^2(x^2)
72k = (1/2)m(2x(dx/dt))
72k = mx(dx/dt)
Now, let's rearrange the equation to solve for dx/dt:
dx/dt = (72k)/(mx) dx/dt = (72k)/(m√(2k/m)x) dx/dt = (72√(k/m))/(√(2k/m)x) dx/dt = (36√2)/x
Now, we have the expression for dx/dt in terms of x. To find the distance from the equilibrium position where kinetic energy and potential energy are equal, we need to solve this differential equation. We can integrate both sides of the equation:
∫(1/x) dx = ∫(36√2) dt
ln|x| = (36√2)t + C
x = e^((36√2)t + C)
Since we are looking for a specific point where kinetic energy and potential energy are equal, we can assume t = 0. This means C = ln(A), where A is the amplitude. Substituting these values, we have:
x = e^((36√2)t + ln(A)) x = Ae^(36√2)t
At this point, we have the equation for the position (x) as a function of time (t). This equation describes the displacement of the mass from the equilibrium position at any given time. To find the distance where kinetic energy and potential energy are equal, we would need to solve for t when the kinetic and potential energies are equal and substitute it back into the equation for x.
However, note that this result is a general expression for the position in the oscillation, and finding the specific distance would require additional information such as the specific time or additional conditions.