Question

A rectangular block of wood floats in a calm lake. Show that, if friction is ignored, when the block is pushed gently down into the water and then released, it will then oscillate with SHM. Also, determine an equation for the force constant.

Short answer

Yes, the block of wood would perform SHM with an equation for the force constant of\(k = {\rho _{{\rm{water}}}}gA\).

Step-by-step solution

Understanding the net force on the block 

The total force on the rectangular block of wood floating in a calm lake will be equivalent to zero in the equilibrium position. So, the buoyancy force is equal to the weight of the block.

Given data

The mass of the block of wood is\(m\).

Block is gently pushed down into the water and then released to create SHM oscillations.

Calculation of buoyancy force

According to the law of equilibrium, all forces must be balanced by the counter forces acting in opposite directions. In an equilibrium position, the block will rest on the water, maintaining both Buoyancy force and weight of the block in opposite directions.

The relation of buoyancy force is given by,

\({F_{{\rm{bouy}}}} = mg\)

Here, g is the gravitational acceleration.

When additional force is applied to the block, then the block will move downwards. However, an equal amount of force of water will push it upwards due to the weight of the additional water displaced. Thus, extra force applied on the block due to extra water is:

\(\begin{aligned}{c}{F_{{\rm{extra}}}} &= {m_{\scriptstyle{\rm{extra}}\atop\scriptstyle{\rm{water}}}}g\\ &= {\rho _{{\rm{water}}}}{V_{\scriptstyle{\rm{extra}}\atop\scriptstyle{\rm{water}}}}g\\ &= \left( {{\rho _{{\rm{water}}}}gA} \right)\Delta x\end{aligned}\)

Here, \({V_{\scriptstyle{\rm{extra}}\atop\scriptstyle{\rm{water}}}}\) is the volume of extra water displaced due to the extra force applied on the block, \({m_{\scriptstyle{\rm{extra}}\atop\scriptstyle{\rm{water}}}}\) is mass of the extra water, \({\rho _{{\rm{water}}}}\) is the density of water, A is the area of the block, and \(\Delta x\) is displacement of the block due to extra force.

Calculation of equation of force constant

Now, to balance the force of extra water applied by the volume of water displaced, an opposite but equal in magnitude force is experienced by the block downwards. Thus, the net force equation applied on the block can be written as:

\({F_{{\rm{net}}}} = - \left( {{\rho _{{\rm{water}}}}gA} \right)\Delta x\)

The ideal equation of motion is given by,

\(F = - k\Delta x\)

Comparing the above two equations, we get the force constant as:

\(k = {\rho _{{\rm{water}}}}gA\)

Hence, the equation of spring constant is \(k = {\rho _{{\rm{water}}}}gA\).