Physics · Oscillations · NEET
Kinetic energy depends on speed, and speed is largest at the mean position (x = 0). So KE is maximum at the centre: K(max) = (1/2)m*w^2*A^2. At the extreme positions (x = A) the body stops for an instant, so v = 0 and KE = 0. In short: fast at the middle means high KE, still at the ends means zero KE.
Start from velocity in SHM: v = w*root(A^2 - x^2). Then K = (1/2)m*v^2 = (1/2)m*w^2*(A^2 - x^2). This form is very useful in NEET because it links KE directly to position. At x = 0 you get the maximum, and at x = A the bracket becomes zero.
KE = (1/2)m*w^2*A^2*cos^2(wt). Using cos^2(theta) = (1 + cos 2*theta)/2, the KE varies as cos(2wt). The angular frequency of the energy is 2w, so the KE graph repeats after time T/2. That is why NCERT says the period of K is T/2. Physically, KE reaches its maximum twice in one full oscillation (once at each pass through the centre).
KE = (1/2)m*w^2*(A^2 - x^2) and PE = (1/2)m*w^2*x^2. Setting them equal gives A^2 - x^2 = x^2, so x = A/root(2). At this point each energy equals half the total: (1/4)m*w^2*A^2. This exact value (A divided by root 2, about 0.707A) is a common NEET answer.
No. KE = (1/2)m*v^2 and v^2 is always positive or zero, so KE is never negative. On a graph the KE curve always stays on or above the time axis. Any option showing KE dipping below zero is wrong at once.
The sum of the kinetic energy and potential energy of a simple pendulum bob is 0.02 J. The speed of the bob at its equilibrium position is approximately (mass of the bob = 20 g)
For a simple pendulum having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by (graph question)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
K = (1/2)m*v^2 = (1/2)m*w^2*(A^2 - x^2) = (1/2)m*w^2*A^2*sin^2(wt + phi). All three forms are equivalent; use whichever fits the given data.
Maximum KE occurs at the mean position (x = 0) and equals (1/2)m*w^2*A^2, which is the same as the total mechanical energy E of the oscillator.
Yes. Total energy E = KE + PE = (1/2)m*w^2*A^2 stays constant throughout the motion. As the body moves, energy shifts between KE and PE, but their sum never changes (no friction).
At the mean position (x = 0), PE = 0, so all the energy is kinetic. There KE = total energy E = (1/2)m*w^2*A^2.
The KE varies at frequency 2n, twice the SHM frequency, because KE depends on cos^2(wt) which has period T/2.
