First slide

Simple hormonic motion

Question

A U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now m gm of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T such that

Moderate

A

$T = 2 π \sqrt{\frac{M}{g}}$
B

$T = 2 π \sqrt{\frac{Ma}{gd}}$
C

$T = 2 π \sqrt{\frac{M}{gdA}}$
D

$T = 2 π \sqrt{\frac{M}{2 Adg}}$

Solution

If the level of liquid is depressed by cm on one side, then the level of liquid in column P is 2y cm higher than B as shown.
The weight of extra liquid on the side P = 2Aydg.
This becomes the restoring force on mass M.
$∴ Restoring acceleration = \frac{- 2 Aydg}{M}$
This relation satisfies the condition of SHM i.e., $a \propto - y$ .
Hence time period
$T = 2 π \sqrt{\frac{Displacement}{Acceleration}}$
= $2 π \sqrt{\frac{\frac{y}{2 Aydg}}{M}} \Rightarrow T = 2 π \sqrt{\frac{M}{2 Adg}}$

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