SHM

JEE Advanced 2015 Papers › Mechanics questions › SHM

A simple pendulum made up of light rod of length L and bob of mass m has a spring of force constant k connected to it at a distance x below the pt. of suspension. Find the frequency of vibration of the system.

#Physics #Mechanics

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6 Answers

1

samagra Kr ·2011-01-28 21:47:04

consider a small ANGULAR displacement of Î¸;
comperession in the spring is>>>> xÎ¸

spring force=kxÎ¸
RESTORING torque abt point of suspension= (kxÎ¸)x=kx²Î¸

torque = MI * Î±

ml²/3(Î±)=-kx²Î¸

:

now you can find it

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6

AKHIL ·2011-01-29 01:45:02

hmm
by this method i m not gettin the rite ans...

i used the energy method and still not gettin the ans!!!

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1

samagra Kr ·2011-01-29 05:49:38

By energy method,

if rod is displaced by Î¸, and have angular velocity Ï‰,

then TOTAL ENERGY=1/2 k(xÎ¸)² +ml²/3 Ï‰²+mgl(1-cosÎ¸)/2 = constant

& 1-cosÎ¸= Î¸²/2 ;

diff wrt Î¸,
Ï‰²=3(kx²+ mgl/2)/2ml²

is it correct now?

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71

Vivek @ Born this Way ·2011-01-29 07:04:37

Nothing very tough in it. If you think clearly you will certainly be able to do it!

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AKHIL ·2011-01-30 22:05:21

ya
i m gettin the same ans by energy method.....
but its wrong!!!

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1

samagra Kr ·2011-01-31 06:21:42

whats the ans?

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