Physics · Oscillations · NEET
At rest, the buoyant force up equals the weight down, so the body floats still. When you push it down by extra depth x, the extra part goes under water, so extra liquid is pushed away. This gives an extra upward buoyant force = (density of liquid) times g times A times x, where A is the cross-section area. This extra force is pushing up, opposite to the downward push, and it grows with x. A force that is always opposite to displacement and proportional to it (F = -kx) means SHM. So the body bobs up and down harmonically.
L is the depth of the part of the body that was already sitting under the liquid when it floated calmly (the submerged length at equilibrium). It is NOT the full height of the body. You can find it from floating condition: weight = buoyant force, so (density of body) times g times (total volume) = (density of liquid) times g times (submerged volume). This gives L = (density of body times total height) / (density of liquid), or simply L = mass / (density of liquid times area A).
Yes. The body's mass and area stay the same, but a denser liquid makes the body sit less deep, so L becomes smaller, and T = 2 pi root(L/g) becomes smaller (faster bobbing). In fact T is proportional to 1 / root(density of liquid). A ReNEET 2026 question uses exactly this: put the same cork in a liquid with double the period, and its density must be one-fourth.
The formula looks identical, T = 2 pi root(L/g), but the meaning of L is different. For a pendulum L is the string length. For a floating body L is the submerged depth. The reason they match is that in both cases the restoring 'stiffness per unit mass' comes out equal to g/L. Do not mix them up in a problem.
For a uniform cylinder of density rho_s, height h, floating in liquid of density rho, the submerged length is L = (rho_s / rho) times h. Substituting into T = 2 pi root(L/g) gives T = 2 pi root(rho_s times h / (rho times g)). So a denser cork (bigger rho_s) sits deeper and oscillates slower; a denser liquid (bigger rho) shortens the period.
A cylindrical cork of uniform density floats in a liquid of density rho1. When depressed slightly and released it oscillates harmonically with time period T. If the same cork floats in another liquid of density rho2, the oscillation has period 2T. The value of rho2/rho1 is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi root(L / g), where L is the depth of the submerged part at equilibrium and g is acceleration due to gravity. L is found from mass = (density of liquid) times L times area.
When pushed down by x, the extra buoyant force = (density of liquid) times g times A times x, directed upward. This is F = -kx with k = (density of liquid) times g times A, which causes SHM.
Slower. A denser body sits deeper, so L is larger, and since T = 2 pi root(L/g), a larger L gives a larger time period.
Yes, for small vertical pushes. The restoring force is proportional to the extra submerged depth and always points back toward equilibrium, so the motion is SHM (as long as the body stays vertical and the cross-section stays uniform).
From the floating condition, L = (density of body / density of liquid) times height of the body. A denser liquid makes L smaller and the period shorter.
