NEET Physics Question (2026) — Solution

Step-by-step solution

The displacement of a simple pendulum executing simple harmonic motion is given by x = A ( t + ). The velocity of the pendulum is v = dx dt = A ( t + ). The kinetic energy is given by K = 1 2 m v^2 = 1 2 m A^2 ^2 ^2( t + ). Since kinetic energy is proportional to the square of velocity, it can never be negative. This eliminates the graphs that show negative values. Also, the kinetic energy of a pendulum is not constant, which eliminates the constant graph. Using the trigonometric identity ^2 = 1 + (2 ) 2 , we get: K = 1 4 m A^2 ^2 [1 + (2 t + 2 )] This shows that the kinetic energy oscillates with an angular frequency 2 , meaning its time period is half that of the pendulum, i.e., T' = T 2 . The graph in option (3) correctly represents a non-negative oscillating quantity with a period of T/2. Answer: