Physics · Oscillations · NEET
Total energy E = KE + PE. Using x = A sin(wt), PE = (1/2)k A^2 sin^2(wt) and KE = (1/2)k A^2 cos^2(wt). Adding them, sin^2 + cos^2 = 1, so E = (1/2)k A^2. This has no x or t in it, so both the E-vs-x graph and the E-vs-t graph are horizontal straight lines at height (1/2)k A^2. Energy just shifts between PE and KE; the total never changes.
At any position x, KE = E - PE = (1/2)k A^2 - (1/2)k x^2 = (1/2)k (A^2 - x^2). So when x grows, PE = (1/2)k x^2 grows (upward parabola) and KE shrinks (downward parabola). They are mirror images across the flat total-energy line. At the mean position (x = 0) KE is maximum and PE is zero; at the extremes (x = +A or -A) PE is maximum and KE is zero.
It depends on the x-axis. Against displacement x, both PE and KE are parabolas because they contain x^2. Against time t, they are sin^2 and cos^2 shaped curves (never straight, never going below zero). Students lose marks by drawing a parabola when the question says 'vs time' or a sine wave when it says 'vs displacement'. Always read the axis first.
KE contains cos^2(wt) and PE contains sin^2(wt). Using cos^2(wt) = (1 + cos 2wt)/2, the cosine inside now has angular frequency 2w. Angular frequency 2w means frequency 2n and period T/2. So the energy curves complete two full humps for every one full oscillation of the particle. This is why a body of frequency n has energy varying at frequency 2n.
Set PE = KE: (1/2)k x^2 = (1/2)k (A^2 - x^2). This gives 2x^2 = A^2, so x = A/root2 (about 0.707 A). At that point each energy is half the total, i.e. (1/4)k A^2. On the graph the two parabolas cross exactly at x = A/root2 on both sides of the mean position.
KE = (1/2)m v^2 uses speed squared, so it can never be negative. PE = (1/2)k x^2 uses x squared and is taken as zero at the mean position, so it is also never negative. Total energy is a fixed positive value. A correct SHM energy graph never dips below the horizontal axis; any option showing a negative dip is wrong.
A body executes simple harmonic motion with frequency n. The frequency of variation of its potential energy is
For a simple pendulum having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by (choose the correct KE-vs-time graph).
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)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
An upward-opening parabola given by PE = (1/2)k x^2, with its lowest point (zero) at the mean position x = 0 and maximum value (1/2)k A^2 at the extremes x = +A and -A.
A downward-opening parabola given by KE = (1/2)k (A^2 - x^2). It is maximum (1/2)k A^2 at x = 0 and drops to zero at x = +A and -A. It is the mirror image of the PE parabola.
Because KE and PE depend on cos^2 and sin^2 of (wt), and squaring a sinusoid doubles its frequency. So if the body oscillates at frequency n, both energies vary at frequency 2n with period T/2.
At x = A/root2 (about 0.707 A). At that position each equals half the total energy, (1/4)k A^2.
No. Total energy E = (1/2)k A^2 = (1/2)m w^2 A^2 is constant. It is a flat horizontal line on both the energy-vs-displacement graph and the energy-vs-time graph, as long as amplitude stays the same.
