Physics · Oscillations · NEET
The liquid gives an upward buoyant force on the bob. This buoyant force acts against gravity, so the net downward pull (the restoring pull) becomes weaker. A weaker pull means the effective gravity g_eff is smaller than g. Since T = 2 pi root(L / g_eff), a smaller g_eff makes T larger. So the pendulum takes MORE time per swing, it does not speed up.
Weight of bob down = d_bob * V * g. Buoyant force up = d_liquid * V * g (weight of displaced liquid). Net downward force = d_bob*V*g - d_liquid*V*g. Divide by the mass d_bob*V to get effective acceleration: g_eff = g(1 - d_liquid/d_bob). The volume V and mass cancel out neatly, leaving only the ratio of the two densities.
No. Both weight and buoyant force are proportional to the same volume V, so V cancels. Only the RATIO d_liquid/d_bob decides g_eff. A denser bob (larger d_bob) makes the ratio small, so buoyancy barely changes T. A bob almost as light as the liquid makes the ratio near 1, so g_eff is tiny and T becomes very large.
Then d_liquid/d_bob = 1, so g_eff = g(1 - 1) = 0. With zero effective gravity there is no restoring force, so the bob simply floats and does not oscillate. The time period becomes infinite (T = 2 pi root(L/0) tends to infinity). This is the limiting case of the formula.
Both change g_eff, but for different reasons. In a liquid, buoyancy subtracts, so g_eff = g(1 - d_liquid/d_bob) and T always increases. In a lift moving up, g_eff = g + a (T decreases); lift moving down, g_eff = g - a (T increases); free fall, g_eff = 0. So the liquid case only ever slows the pendulum, never speeds it up.
A simple pendulum oscillating in air has a period of root(3) s. If it is completely immersed in a non-viscous liquid whose density is one-fourth that of the material of the bob, the new period is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi root(L / g_eff), where g_eff = g(1 - d_liquid/d_bob). Because g_eff is less than g, the period is longer than the normal period in air.
It increases the time period. Buoyancy reduces the effective gravity, and a smaller g always makes T larger, so the pendulum swings slower.
Viscosity adds a damping (drag) force. The pendulum still slows down due to buoyancy AND its amplitude keeps shrinking with time (damped oscillation). NEET problems usually say 'non-viscous' so only buoyancy matters and amplitude stays constant.
Both the weight (d_bob * V * g) and the buoyant force (d_liquid * V * g) contain the same volume V. When you compute net force per unit mass, V and the constants cancel, leaving g_eff = g(1 - d_liquid/d_bob), which depends only on the density ratio.
Yes, if the bob density equals the liquid density, g_eff becomes zero. Then there is no restoring force and the bob just floats, so the time period becomes infinite and it does not oscillate.
