Physics · Oscillations · NEET
Suppose the liquid on the left arm drops by a small distance y. Then the liquid on the right arm rises by the same y (the liquid is not compressed). Now one side is higher than the other by a height difference of 2y. This extra column of height 2y has weight, and that weight acts as a restoring force pulling the liquid back. The restoring force comes out proportional to the displacement y and points back toward the balanced position. Force proportional to displacement, directed opposite to it, is exactly the condition for SHM.
Let A be the cross-section area, rho the density, and L the total length of liquid. Extra height difference = 2y, so extra weight = (2y)(A)(rho)(g). This is the restoring force: F = -2 rho A g y. Total mass of moving liquid m = rho A L. Using F = m a: -2 rho A g y = (rho A L) a, so a = -(2g/L) y. Compare with SHM a = -(omega squared) y: omega squared = 2g/L. Then T = 2 pi / omega = 2 pi root(L / 2g). The area A and density rho cancel out.
No. The density rho appears in both the restoring force and the moving mass, so it cancels completely. Whether you use water, oil, or mercury, the same length of column in the same tube gives the same time period. This is a favourite trap - the answer does not depend on the liquid used.
L is the TOTAL length of the liquid column, measured along the whole path of the liquid (down one arm, across the bend, up the other arm). It is not the length of a single arm and not the height. If the total liquid length is L, then T = 2 pi root(L/2g).
When the liquid drops by y on one side, it rises by y on the other side. So the unbalanced height that creates the restoring force is 2y, not y. That factor of 2 stays in the restoring force and lands in the formula as the 2 in L/2g. If you forget it and write T = 2 pi root(L/g), you will get a wrong (too large) answer.
A simple pendulum has T = 2 pi root(l/g). The U-tube gives T = 2 pi root(L/2g) = 2 pi root((L/2)/g). So the liquid column behaves exactly like a pendulum whose length equals L/2, that is, half the total length of the liquid column. Same math, different setup.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi root(L / 2g), where L is the total length of the liquid column and g is the acceleration due to gravity. The angular frequency is omega = root(2g/L).
No. The density appears in both the restoring force and the moving mass and cancels out. Water, oil, and mercury all give the same period for the same column length in the same tube.
No. The area A also cancels in the derivation. A wide or narrow uniform tube gives the same time period for the same total length L.
L is the total length of the liquid, measured along its full path through both arms and the bend. Not one arm, and not the vertical height.
They share the same form. T = 2 pi root(L/2g) is a pendulum of effective length L/2. So the liquid column swings like a pendulum half as long as the total column.
When the liquid drops by y on one side and rises by y on the other, the height difference is 2y. The weight of this extra 2y column pulls the liquid back toward balance, giving F = -2 rho A g y.
