Home
Class 11
PHYSICS
A cylindrical piece of cork of density o...

A cylindrical piece of cork of density of base area Aand height h floats in a liquid of density `rho_l`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T=`2pisqrt((hrho)/(rho_1g))` where p is the density of cork. (Ignore damping due to viscosity of the liquid)

Text Solution

Verified by Experts

Let l be height of cork. At equilibrium weight of the cork = upthrusts of the portion of the cork inside the liquid
`Ah rhog=Alrho_(L)g`
`l=(h rho)/(rho_(L))` . . . . (i)
If the cork is slightly depressed below through a distance y. then net upward force is
`F=-[A(l+y)rho_(L)g-Al rho_(L)g)`
`=-A rho_(L)gy`
Negative sigh shows that force is opposite to displacement.
If a be the acceleration produced, then `a=(F)/(m)`
`a=-(A rho_(L)gy)/(A rhoh)` . . . (ii) `(because m=Arho h)`
The above expression represents equation of motion.
Comparing equation (ii) with the equation
`a=-omega^(2)y`
`:.` We get, `omega^(2)=(rho_(L)g)/(h rho)" ":." "omega=sqrt((rho_(L)g)/(h rho))`
`:.` Time period of oscillation is `T=(2pi)/(omega)=(2pi)/(sqrt((rho_(L)g)/(hrho)))rArrT=2pisqrt((hrho)/(rho_(L)g))`
Promotional Banner

Similar Questions

Explore conceptually related problems

A cylindrical log of wood of height h and area of cross-section A floasts in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period T = 2pi sqrt(m//Arhog) where m is mass of the body and rho is density of the liquid.

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called 'law of atmospheres' n_2 = n_1 exp[-(mg)(H_2-H_1)/(kT)] where n_2,n_1 refer to number density at heights h_2 and h_1 respectively.Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_2= n_1 exp[(-mgN_a)/(RT) (1-(rho)/(rho))(h_2-h_1)] where rho is the density of the suspended particle, and rho' that of surrounding medium. [N_A is Avogadro's number, and R the universal gas constant].