OscillationsNEET Physics · Class 11 · NCERT Chapter 13

You can expect 1 to 2 questions from Oscillations in NEET 2027. The chapter has high PYQ frequency. Time period of pendulum and spring-mass, energy in SHM, velocity and acceleration at given displacement, and series / parallel spring problems are the favourites.

SHM is a special kind of oscillatory motion in which the restoring force is directly proportional to the displacement from a fixed point and is always directed toward that point. Mathematically, F equals minus k x, which gives a equals minus omega squared x, with the standard solution x of t equals A sin(omega t plus phi).

For small angles, T equals 2 pi times the square root of L over g, where L is the length of the pendulum and g is the acceleration due to gravity. The period does not depend on the mass of the bob or the (small) amplitude. NEET problems often vary L or g (different planets, lifts) and ask how T changes.

T equals 2 pi times the square root of m over k, where m is the mass and k is the spring constant. This is independent of the amplitude. The period is the same on a horizontal frictionless surface and hanging vertically; gravity only shifts the equilibrium position, it does not change T.

For x equals A sin(omega t plus phi), v equals A omega cos(omega t plus phi) and a equals minus A omega squared sin(omega t plus phi). At the mean position (x = 0), v is maximum equal to A omega and a is zero. At the extremes (x = plus minus A), v is zero and a is maximum equal to A omega squared. A useful formula: v equals omega times square root of (A squared minus x squared).

The total mechanical energy of SHM is constant: E equals half k A squared equals half m omega squared A squared. KE equals half k (A squared minus x squared) and PE equals half k x squared. KE is maximum at x equals 0; PE is maximum at x equals plus minus A. Their sum is always E.

Series: the same force stretches each spring, so they extend in turn. Effective spring constant is given by 1 over k_eff equals 1 over k_1 plus 1 over k_2. The effective k is smaller than the smallest k. Parallel: both springs share the load and extend by the same amount. k_eff equals k_1 plus k_2. The effective k is larger than the largest k.