Question

Consider a wooden block in the shape of a cube whose edges are 10 cm long. The density of the wood is 0.8 g /cm² . The block is submersed in water; a guiding mechanism guarantees that the top and the bottom surfaces of the block are parallel to the surface of the water at all times. Let x(t)be the depth of the block in the water at time t. Assume that xis between 0 and 10 at all times.

a.Two forces are acting on the block: its weight and the buoyancy (the weight of the displaced water).

Recall that the density of water is 1 g/cm ³. Find formulas for these two forces.

b.Set up a differential equation for x(t). Find the solution, assuming that the block is initially completely submersed [x(0)=10] and at rest.

c.How does the period of the oscillation change if you change the dimensions of the block? (Consider a larger or smaller cube.) What if the wood has a different density or if the initial state is different? What if you conduct the experiment on the moon?

Short answer

1. $100x\left(t\right)\times 980=98000x\left(t\right)$gram
2. $x\left(t\right)=2\cos \left(\sqrt{122.5t}\right)+b\sin \left(\sqrt{122.5t}+8\right)$
   
3. $V={\left(10\right)}^3=1000c{m}^3$
   

Step-by-step solution

Consider for part (a).

First consider the cube with a=10cm and density 0.8 g/cm³

So we have

$V={\left(10\right)}^3=1000c{m}^3$

It holds

Mass=density*volume =0.8*100=800grmas

So, we can conclude that the weight of the cube is

$mg=800\times 980=784000$

Since, we know that the value of x(t) is the depth of block in water at time t.

$x\left(t\right)\times 10\times 10=100x\left(t\right)$

Then the mass is,

$100x\left(t\right)\times 1gram$

And the weight if displaced water is$100x\left(t\right)\times 980=98000x\left(t\right)$ gram

Consider for part (b).

Let assume that initially the box is at rest or x(0)=10

The equation of motion is

$$\begin{gathered} mx\text{'}\text{'}\left(t\right)=mg \\ m=10\times 10\times 10\times 0.8=800grams \end{gathered}$$

And so,

$$\begin{gathered} 800x\text{'}\text{'}\left(t\right)=800\times 980-98000x\left(t\right) \\ 10x\text{'}\text{'}\left(t\right)=9800-1225x\left(t\right) \\ 10x\text{'}\text{'}\left(t\right)+122.5x=980 \end{gathered}$$

So the final equation is

$$\begin{gathered} D^2+122.5=0 \\ D=\pm \sqrt{122.5i} \end{gathered}$$

So, we have now

$$\begin{gathered} x\left(t\right)=a\cos \left(\sqrt{122.5t}\right)+b\sin \left(\sqrt{122.5t}\right)\frac{980}{122.5} \\ x\left(t\right)=a\cos \left(\sqrt{122.5t}\right)+b\sin \left(\sqrt{122.5t}\right)+8 \\ x\left(0\right)=10 \end{gathered}$$

So, we have,

$$\begin{gathered} a+8=10 \\ a=2 \end{gathered}$$

So, the final equation is$x\left(t\right)=2\cos \left(\sqrt{122.5t}\right)+b\sin \left(\sqrt{245t}+8\right)$

Consider for part (c).

In the case when we have the cube of side 5cm, the equation of motion is

$$\begin{gathered} x\text{'}\text{'}\left(t\right)+245x\left(t\right)=4900 \\ x\left(t\right)=a\cos \left(\sqrt{245t}\right)+b\sin \left(\sqrt{245t}+20\right) \end{gathered}$$

So, the period the oscillation decrease because we know the frequency of the oscillation increases.

In the same way, the period of the oscillation increases when we have a large cube.

Hence, First consider the cube with a=10cm and density 0.8 g/cm³

So the values of V is $V={\left(10\right)}^3=1000c{m}^3$.