A U-tube of uniform cross-sectional area A = 1 × 1 0 − 3 m 2 contains a liquid of density ρ = 1000 kg/m 3 . The total length of the liquid column is L = 6.5 m. A piston of mass m = 1 kg is placed on t

Question

A U-tube of uniform cross-sectional area A = 1 × 1 0 − 3 m 2 contains a liquid of density ρ = 1000 kg/m 3 . The total length of the liquid column is L = 6.5 m. A piston of mass m = 1 kg is placed on the liquid surface in one arm of the U-tube. A vertical spring of spring constant k = 10 N/m is attached to the piston, with its other end fixed such that the spring is in its natural length when the system is in equilibrium. If the piston is slightly displaced from its equilibrium position and released, the system performs simple harmonic motion. Assuming g = 10 m/s 2 and neglecting friction and viscosity, find the value of π 4 T ​ , where T is the time period of oscillation in seconds.

TestHub · Accepted Answer

The total mass oscillating is the mass of the piston plus the mass of the liquid column: $M_{t o t a l} = m + ρ A L = 1 + (1000) (1 \times 1 0^{- 3}) (6.5) = 1 + 6.5 = 7.5$ kg.

The effective restoring force constant combines the spring constant and the hydrostatic restoring force. When the piston moves down by $x$ , the liquid level in that arm drops by $x$ and rises by $x$ in the other arm, creating a height difference of $2 x$ . The restoring force due to gravity on the liquid is $F_{g} = (2 x) ρ A g$ . The restoring force from the spring is $F_{s} = k x$ . Thus, the effective spring constant is $K_{e ff} = 2 ρ A g + k = 2 (1000) (1 \times 1 0^{- 3}) (10) + 10 = 20 + 10 = 30$ N/m. The time period of oscillation is $T = 2 π \frac{M _{t o t a l}}{K _{e ff}} = 2 π \frac{7.5}{30} = 2 π \frac{1}{4} = 2 π (\frac{1}{2}) = π$ seconds. The value of $\frac{4 T}{π} = \frac{4 π}{π} = 4$ .

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