1.

A cylindrical block of wood of mass M is floating n water with its axis vertica. It is depressed a little and then released. Show that the motion of the block is simple harmonic and find its frequency.

Answer» Suppose a height `h` of the block is dipped in the water in equilibrium position. If `r` be the radius of the cylindrical block, the volume of the water displaced `= pir^(2)h`. For floating in equilibrium,
`pir^(2)hrhog = W` ……..(i)
where `rho` is the density of water and `W` the weight of the block.
Now suppose during the vertical motion, the block is further dipped through a distance `x` at some Instant. The volume of the displaced water `pir^(2)(h + x) rhog` vertically upward.
Net force on the block at displacement `xx` from the equilibrium position is
`F = W - pir^(2)(h + x)rhog = W - pir^(2)hrhog - pir^(2)rho xx g`
Using `(i) F = -pir^(2) rhogx = -kx`,
where `k = pir^(2)rhog`.
Thus, the block executes `SHM` with frequency.
`v = (1)/(2pi) sqrt((k)/(m)) = (1)/(2pi) sqrt((pir^(2)rhog)/(m))`


Discussion

No Comment Found

Related InterviewSolutions