A classical mechanics problem by Pratik Shastri

Classical Mechanics Level 4

A pendulum oscillates, immersed in water. The mass of the bob is $250 \ g$ , length of the string is $30 \ cm$ , the radius of the bob is $2 \ cm$ and density of water $\rho_w$ is $1000 \ kg/m^3$ .

Calculate the time period of its oscillations (in seconds), considering its amplitude to be small.

$\textbf{Assumptions}$

$\bullet$ Take $g=9.8 \ m/s^2$

$\bullet$ Ignore drag forces.

The answer is 1.18135.

2 solutions

Ronak Agarwal
Aug 5, 2014

Time period of a pendulum = $2\pi \sqrt { \frac { l }{ { g }_{ eff } } }$

Also ${ g }_{ eff }=\frac { Net\quad force }{ Net\quad mass } =\frac { Weight-Buoyant\quad force }{ mass } =\frac { { { \rho } }_{ 1 }Vg-{ \rho }_{ 2 }Vg }{ { \rho }_{ 1 }Vg }$

$\Rightarrow { g }_{ eff }=\frac { { \rho }_{ 1 }-{ \rho }_{ 2 } }{ { \rho }_{ 1 } }$

${ \rho }_{ 1 }=\frac { M }{ V } =\frac { 0.25\quad Kg }{ \frac { 4 }{ 3 } \pi { (0.02) }^{ 3 }\quad { m }^{ 3 } } =7460\quad Kg/{ m }^{ 3 }$

Put the values to get $\boxed{T=1.181 s.}$

Sir, maybe you've made a typo. In the Second Line, net mass will be (Ro1)(V). Therefore, g(eff.)=(Ro1-Ro2)(g)/(Ro1). Moreover, The Ro1 is exactly, (7460.387957 Kg per meter cube). If we calculate the Time Period by using this EXACT Value(Not the approximate 7460), then we get the Time period (1.22009 sec). This is actually the Exact Time Period. Shouldn't we use Exact Value Instead of Approximate 7460 ? Kindly notice this, Sir. I posted the Exact Value and I was given wrong! Why? Why should we use Approximate values when we can use the Exact ones?

Rubayet Tusher - 6 years, 1 month ago

Did exactly the same way !!

CH Nikhil - 6 years, 1 month ago

Andrew Song
Sep 4, 2015

A classical mechanics problem by Pratik Shastri

The answer is 1.18135.

2 solutions

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