Question 292: Oscillations

Question

A tube of uniform bore of cross-sectional area ' $A$ ' has been set up vertically with open end facing up. Now ' $M$ ' gram of a liquid of density ' $d$ ' is poured into it. The column of liquid in this tube will oscillate with a period ' T ', which is equal to [ $g=$ acceleration due to gravity]

Accepted Answer

For a depression of y cm on one side, the level of liquid will be 2 y cm higher on the other side.
$	herefore \quad$ Weight of extra liquid on the P -side $=2$ Aydg The above weight acts as the restoring force acting on mass M .
$	herefore \quad$ Restoring acceleration $=-\frac{2 \mathrm{Aydg}}{\mathrm{M}} \ldots(-\mathrm{ve}$ sign indicates the acceleration and force are opposite to each other) ... (i)
We know for simple harmonic motion, $a=-\left(\frac{k}{m}ight) x\quad	ext{.... (ii)}$
Comparing (i) and (ii), we see the motion is simple harmonic.
$$\begin{aligned}
	herefore \quad \mathrm{T} & =2 \pi \sqrt{\frac{	ext { Displacement }}{	ext { Acceleration }}}=2 \pi \sqrt{\frac{\mathrm{y}}{\frac{2 \mathrm{Aydg}}{\mathrm{M}}}} \
& =2 \pi \sqrt{\frac{\mathrm{M}}{2 \mathrm{Adg}}}
\end{aligned}$$

Answer

$2 \pi \sqrt{\frac{\mathrm{MA}}{\mathrm{gd}}}$

Answer

$2 \pi \sqrt{\frac{\mathrm{M}}{2 \mathrm{Adg}}}$

Answer

$2 \pi \sqrt{\frac{M}{g}}$

Answer

$2 \pi \sqrt{\frac{M}{g d A}}$

Oscillations

A tube of uniform bore of cross-sectional area ' A ' has been set up vertically with open end facing up. Now ' M ' gram of a liquid of density ' d ' is poured into it. The column of liquid in this tube will oscillate with a period ' T ', which is equal to [ g= acceleration due to gravity]

Choose the correct answer: