First slide

Applications of SHM

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 y 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 $P = 2 Aydg$ .
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 $T = 2 π \sqrt{\frac{Displacement}{|Acceleration|}}$
$= 2 π \sqrt{\frac{y}{\frac{2 Aydg}{M}}} \Rightarrow T = 2 π \sqrt{\frac{M}{2 Adg}}$

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