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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. |
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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))` |
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