no one for this q^{n} ?
A body of mass 200 grams is in equilibrium At x equal to zero under the influence of a force f(x) =( 100x+10x^{2} )N.
( A)If is the body is displaced a small distance from equillibrium what is the period of its association?
(B) if the amplitude is 4. 0 centimetre, by how much do we in assuming that f(x)= kx at the end point of the motionl

UP 0 DOWN 0 1 4
4 Answers
A)
kx=10x^{2}100x
or, k=10x100
since, the body is displaced by a small distance so we can neglect 10x
therefore,
k=100
now, we know that k=Ï‰^{2}m
therefore,
Ï‰^{2}m=100
or,Ï‰=10âˆš5
now,
T=2Ï€/Ï‰
or,T=Ï€/5âˆš5
B)
actual force=10x^{2}100x=(16/1000)4
assumed force=kx=4
therefore,
error in calculation=[(16/1000)4(4)/(416/1000)]100
=0.4%
 Himanshu Giria thanx
Upvote·0· Reply ·20140201 02:21:54
(B) if the amplitude isÂ 4. 0 centimetre, by how much do we error
in assuming that f(x)= kx at the end point of the motion
Time period = Ï€/5âˆš5 and
error in calculation = 0.4%
 Himanshu Giria the ans is right . plz show how ..
 Akash Anand Nandikesh...I guess u have asked this question to me. Please post the complete solution.
 Akash Anand Time period will be pi/5(sqrt2)
 Himanshu Giria yes sir that is the ans but pl show how ...