Density of cylinder that undergoes vertical oscillation

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To determine the mass density of a metal cylinder undergoing vertical oscillation in mercury, a diagram illustrating the cylinder's displacement from equilibrium is essential. The acceleration of the cylinder can be expressed using relevant physics equations, incorporating the known density of mercury, which is 13600 kg/m^3. The period of oscillation, given as 0.6 seconds, is also crucial for calculations. By applying principles of buoyancy and oscillatory motion, the mass density of the cylinder can be derived. Understanding these foundational concepts is key to solving the problem effectively.
dobbygenius
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Homework Statement
A metal cylinder with a height of 14 cm floats in mercury so that it undergoes a vertical oscillation with a period of 0.6 s. What is the mass density of the cylinder?
Relevant Equations
density of mercury: 13600 kg/m^3
I'm not sure where to start...
 
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dobbygenius said:
Homework Statement:: A metal cylinder with a height of 14 cm floats in mercury so that it undergoes a vertical oscillation with a period of 0.6 s. What is the mass density of the cylinder?
Relevant Equations:: density of mercury: 13600 kg/m^3

I'm not sure where to start...
Start with a diagram showing the cylinder at some displacement from equilibrium.
Next, write an equation for its acceleration.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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