Question

Straight wooden rod of mass 6.0 kg, uniform cross-section 7 cm², and constant density 700 kg has a very small mass 0.8 kg attached to its one end. The rod is partially submerged in water of density 990kg/m³.  While in equilibrium, the rod floats in a vertical position with large part of it submerged. The rod is then pushed down by distance y_max from equilibrium and released resulting in its oscillation neat the surface of teh water.

(Assuming no drag forces are acting in this situation) and taking g to be 9.8 m/s²

A) Use the first principles to demonstrate that such system can be treated as Simple Harmonic Oscillator. ( Paper solution!!!)

B) Find the rod's period of oscillations to the nearest thousandth of a second.

( Enter this value into the answer box without units).

In addition to entering your final numerical answer into the box, make sure that you write your solution neatly starting with the clear diagram and all variables on it.

Solve this problem in detail on paper. Please annotate your solution -- make short comments/ arguments for steps you are making.

Answer 1

Let mass of the rod be , mass of the small mass be , length of the rod be , density of the rod be , density of water be , area of the rod be and the length of the rod that is immersed in water when its in equilibrium be

1. equilibrium position

force of bouyency

force of gravity

2. displaced by a small distance

force of bouyency

force of gravity

net force = force of bouyency -force of gravity

from newtons second law of motion

net force

negative sign indicates the direcition is opposite.

from the above you can see acceleration is directly proportional to displacement but in opposite direction.

we know that this behavior implies the motion is 'simple harmonic motion'

in simple harmonic motion

time period of oscillation