A rectangular tank (base 15 cm × 20 cm) contains water of density 1000 kg m⁻³ to height 20 cm. A vertical spring of natural length h₀=20 cm and constant K=280 N m⁻¹ is fixed to the bottom. The system is in an elevator accelerating downward at 2 m s⁻². A cube of side 10 cm and mass 2 kg is placed on the spring. If its vertical oscillation frequency is a√2/π, find a.

Question

G1 Physics · Accepted Answer

Tank base area=300 cm² and cube area=100 cm², so when the cube moves down by x, water rises by y=100x/(300-100)=x/2. Hence extra submerged depth is x+y=3x/2. In the downward-accelerating lift, g′=8 m s⁻², so buoyancy contributes kᵦ=ρg′ A(3/2)=120 N m⁻¹. Total kₑ=280+120=400 N m⁻¹. With m=2 kg, f=(1/2π)√(400/2)=5√2/π, so a=5.