what is the kinetic energy of a body executing simple harmonic motion is maximum at ???
Answers
At each point in the oscillation, it is the sum of the kinetic energy of the mass plus the potential energy of the spring. It turns out that the sum of these two is a constant, as I’ll demonstrate.
The kinetic energy of the moving mass is 1/2 m V^2
The potential energy of the stretched spring is 1/2 k x^2
Suppose the motion is x = A sin (omega t)
For a spring mass system, omega = squareroot (k/m)
The potential energy at each x is therefore
PE(x) = 1/2 k x^2 = 1/2 k A^2 sin^2(omega t)
The velocity is the derivative of the position
V = A omega cos (omega t)
KE = 1/2 m V^2 = 1/2 m A^2 omega^2 cos^2(omega t)
= 1/2 m A^2 [sqrt(k/m)]^2 cos^2(omega t)
= 1/2 m A^2 k/m cos^2 (omega t) = 1/2 A^2 k cos^2 (omega t)
So the sum PE and KE is
PE + KE = 1/2 k A^2 sin^2(omega t) + 1/2 k cos^2 (omega t)
= 1/2 k A^2 [sin^2(omega t) + cos^2(omega t)]
=1/2 k A^2
This is a constant independent of x. So the total energy is the same at all points in the cycle. This particular value that we found happens to be the potential energy when x = A, the maximum amplitude, at which point V = 0.
We could also have substituted differently and ended up with 1/2 m Vmax^2.
Both expression have the same numerical value. They are equal.