Simple Harmonic (5).

A thin uniform disc of mass M M and radius R R is hinged at its lowest point and placed in a gravity-free space. It is attached by two identical and ideal springs of force constants k k each and a string of cross-sectional area A A , length \ell and Young's modulus Y Y at the positions shown in the figure (centre, top and top). The springs are currently in their natural lengths, the string is unstretched and the whole system is in equilibrium. The disc is given a small angular displacement about the hinge P P and allowed to oscillate freely. If the time period of the small oscillations can be expressed as

T = π ( a M b k + a M b k + c A Y ) T = \pi \left( \sqrt{\dfrac{aM}{bk}} + \sqrt{\dfrac{aM\ell}{bk\ell + cAY}} \right)

where ( a , b ) (a,b) and ( a , c ) (a,c) are pairwise coprime positive integers and a a and b b are both square free. Find a + b + c a+b+c .


The answer is 21.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Atman Kar
Apr 27, 2018

Correct approach! Liked the solution. :)

Tapas Mazumdar - 3 years, 1 month ago

Why the string loses tension?

Alapan Das - 2 years, 2 months ago

Log in to reply

It starts slacking.

Atman Kar - 2 years ago
Sahil Silare
Apr 18, 2018

a = 3 , b = 10 a=3 , b=10 and c = 8 c=8

I am not quite sure, But I got 2 π 2 \pi instead of π \pi before the bracket. Can you check or should I recheck my calculations?

Md Zuhair - 3 years, 1 month ago

Log in to reply

I've confirmed with Tapas after the day I solved.

Sahil Silare - 3 years, 1 month ago

Log in to reply

Okay. I see.... Then lemme check once again

Md Zuhair - 3 years, 1 month ago

You must've considered two separate oscillations independently (which is correct) but notice that each oscillation contributes to only half the time period of that particular oscillation. Hence the term outside the brackets is π \pi . If there had been no string there, the term c A Y cAY equates to zero and you get the time period as a standard result with 2 π 2\pi term outside.

Tapas Mazumdar - 3 years, 1 month ago

Log in to reply

Oh yes!! Thanks! I now got it!

Md Zuhair - 3 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...