Question

A cubic block of side \(l\) and mass density \(\rho\) per unit volume is floating in a fluid of mass density \(\rho_{0}\) per unit volume, where \(\rho_{0}>\rho .\) If the block is slightly depressed and then released, it oscillates in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion. Hint Use archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (bouyant) equal to the weight of the displaced fluid.

Short answer

Answer: The period of the oscillatory motion is 2π.

Step-by-step solution

Find the volume submerged when the block is in equilibrium

When the block is in equilibrium, the buoyant force acting on the object will be equal to its weight. Let \(V_s\) be the volume of the block submerged in the fluid. Using Archimedes' principle, \[\rho_{0}gV_s = \rho gl^3\] Solving for \(V_s\), \[V_s = \frac{\rho}{\rho_0}l^3\]

Determine the additional submerged volume when the block is depressed

Let the block be depressed by a distance \(y\) further into the fluid. Then, the additional volume \(V_{add}\) submerged is given by, \[V_{add} = l^2y\]

Establish the relationship between buoyant force and new submerged volume

Due to the additional submerged volume, the buoyant force will increase. Using Archimedes' principle again for the total submerged volume \((V_s + V_{add})\), we have \[\rho_{0}g(V_s + V_{add}) = (\rho_{0}gV_s) + (\rho_{0}gl^2y)\]

Derive the differential equation of motion for the block

We can now use Newton's second law to find the equation of motion. The net force acting on the block is given by the difference between the buoyant force and the weight of the block. The acceleration of the block is given by \(\frac{d^2y}{dt^2}\). \[\rho_{0}gl^2\frac{d^2y}{dt^2} = (\rho_{0}gV_s) + (\rho_{0}gl^2y) - (\rho gl^3)\] Plugging the expression for \(V_s\) from step 1, \[\rho_{0}gl^2\frac{d^2y}{dt^2} = (\rho_{0}g\frac{\rho}{\rho_0}l^3) + (\rho_{0}gl^2y) - (\rho gl^3)\] Simplifying gives, \[\rho_{0}l^2\frac{d^2y}{dt^2} = \rho_{0}l^2y\] Dividing by \(\rho_{0}l^2\), \[\frac{d^2y}{dt^2} = y\] This is the differential equation of motion for the block.

Find the period of oscillatory motion

The equation we derived in step 4 is a simple harmonic oscillator equation, with the general form \[\frac{d^2y}{dt^2} = -\omega^2y\] where \(\omega\) denotes the angular frequency. Comparing with our previous equation, we have \[\omega^2 = 1\] Thus, \(\omega = 1\) (we can disregard the negative solution here, as frequency must be positive). The period (T) of the oscillatory motion is given by \[T = \frac{2\pi}{\omega}\] Plugging in our calculated value for \(\omega\), \[T = 2\pi\] Hence, the period of the oscillatory motion is \(2\pi\).