Q.
A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On slightly pressing it downwards and releasing, it starts executing simple harmonic motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is close to (densityofwaterρ=103kg/m3)
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a
3.9 rad/s
b
12.7 rad/s
c
153 rad/s
d
127 rad/s
answer is B.
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Detailed Solution
The restoring force responsible for executing simple harmonic motion in this case is the buoyant force.Using this, we can equate FB=ρVg to the restoring force F=kx .Comparing the two equations, kx=ρVgHere V=Ax,Substituting this we get kx=ρAxgHence, k=ρAg For a body executing simple harmonic motion, its angular frequency can be written as ω=kmSubstituting k=ρAg in ω=km, we get ω=ρAgm .Now, the mass of water in the bottle m=ρAhHence, ω=ρAgρAh=gh………………………(1)In the question, the volume of water is given as 500 ml. Using the conversion that 1l=10−3 m3, we get 1ml=10−6 m3,Hence, volume of 500ml of water is 5×10−4 m3Again using V=Ah, 5×10−4 =AhTherefore h=5×10−4 A=5×10−4 πr2. Given the radius if the base is 5 cm,h=5×10−4 π×5×10−22 = 1 5πh=115.70 .Substituting this in equation (1), ω=gh=9.8115.70ω=12.7 rad/s
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