Energy In Shm

Question

A mass oscillating in simple harmonic motion has total mechanical energy E, and at a certain instant its displacement equals half the amplitude. What fraction of the total energy is then in the form of kinetic energy?

RankGuru · Accepted Answer

In SHM the total energy stays constant at $E = \frac{1}{2}kA^2$, continuously trading between potential and kinetic forms. The kinetic energy can be written as $K = \frac{1}{2}k(A^2 - x^2)$, which shows the energy of motion shrinks steadily as the displacement grows toward the amplitude. The potential energy at displacement $x$ is $U = \frac{1}{2}kx^2$, so at $x = A/2$ we get $U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2ight) = \frac{1}{4}E$. By conservation of energy the kinetic energy is the remainder, $K = E - \cup = E - \frac{1}{4}E = \frac{3}{4}E$. The option $\frac{1}{4}E$ actually gives the potential energy and confuses the two stores. The option $\frac{1}{2}E$ would only hold at a special displacement of $A/\sqrt{2}$, not at $A/2$. The option $\frac{1}{3}E$ has no consistent basis in the quadratic energy relations. This uses the NCERT energy treatment of SHM where potential energy scales as the square of displacement. As a sanity check, since the body is still well inside its turning point at $x = A/2$, it is moving fast and should carry most of the energy as kinetic, consistent with the dominant $\frac{3}{4}E$ fraction.

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