One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Explanation

(i) Area of cross-section of the U-tube = A
Density of the mercury column =p
Acceleration due to gravity = g
Restoring force, F = Weight of the mercury column of a certain heightWhere, 2h is the height of the mercury column in the two armsk is a constant, given by Time period, Where, m is the mass of the mercury columnLet l be the length of the total mercury in the U-tubeMass of mercury, Volume of mercury × Density of mercury = Hence, the mercury column executes simple harmonic motion with time period (ii)Volume of the air chamber = VArea of cross-section of the neck = aMass of the ball = mThe pressure inside the chamber is equal to the atmospheric pressure.Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume of the air chamber, DeltaV=ax
Volume strain=`("Change in volume ")/("Original volume")`
`DeltaV/V= (ax)/V`
bulks Modulus of air, B=`("Stress")/("Strain")=(-P)/(ax)//V`
In this case, In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
`p=(-Bax)/V`
The restoring force acting on the ball,
`F=pxxa=(-Bax)/V`
`=(-Ba^(2)x)/V`.......(i)
In simple harmonic motion, the equation for restoring force is:
F=-kx.....(ii)
Where ,k is the spring consatnt
Comparing equation (i) and(ii), we get `k=(Ba^(2))/V` Time period , `T=2pisqrt(m/k)=2pisqrt((vm)/(Ba^(2))`