Saved Bookmarks
| 1. |
A uniform cylinder of length (L) and mass (M) having cross sectional area (A) is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half - submerged in a liquid of density (rho) at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is (k), the prequency of oscillation of the cylindcer is. |
|
Answer» As the cylinder is half submerged in a liquid of density` rho` so upthrust on cylinder `=A(L//2)rhog` . If l is the extension of spring in equilibrium position, then `kl=Mg-A(L//2)rhog` ..(i) If y is the displacement of cylinder from equilibrium position downwards, then restoring force `F=[k(l+y)-{Mg-A(L//2+y)rhog}]` `-[Mg-(AL)/(2)rhog+ky-Mg+A((L)/(2)+y)rhog]` `[`From (i) `]` `=-[ky+Ayrhog)=-(k+Arhog)y` ...(ii) From (ii), `Fpropy` and `-ve` sign show that the force F is directed towards the mean position, hence the cylinder if left free will execute SHM. Comparing (ii) with the relatin, `F=-Ky,` we have spring factor, `K=(k+Arhog)` Inertia factor `=` mass of cylinder `=` M Frequency of oscillation, `v=(1)/(2pi)sqrt((spri ng fact o r)/(i n e rtia fact o r))` or `v=(1)/(2pi)sqrt((k+Arhog)/(M))` |
|