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December 18, 2014

December 18, 2014

Posted by **Jay** on Sunday, July 18, 2010 at 12:26am.

a) Its mass is doubled. What is its period now?

b) Its length is doubled. What is its period now?

The original pendulum is taken to a planet where g = 16 m/s2.

c) What is its period on that planet?

- College Physics. -
**drwls**, Sunday, July 18, 2010 at 1:34amYou should be able to figure these out yourself using the formula for the Period:

P = 2 pi*sqrt(L/g)

Note that it does not depend upon the mass of the pendulum.

- College Physics. -
**Jay**, Sunday, July 18, 2010 at 2:44amreally?? but how would i find L?

- College Physics. -
**drwls**, Sunday, July 18, 2010 at 6:09amIf you double L (the length), P increases by a factor sqrt2 = 1.414, no matter what L is.

- College Physics. -
**manoj**, Saturday, July 31, 2010 at 2:37am(a) time period is independent of mass. so time period of pendulum remains same.

(b)

let initial time period be t1 when length is l and t2 be time period after double of length,i,then

t1=sqrt(l/g)....(1)

t2=sqrt(2l/g)

=root 2xsqrt(l/g)....(2)

from 1 and 2,

t2=root 2 times t1

where t1=1.8 s

(c)ti/t2=sqrt(g'/g)

where g=9.8m/s,ti=1.8s and g'=16m/s

solve n u will get answer

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