A body of mass 200 grams is in equilibrium At x equal to zero under the influence of a force f(x) =( -100x+10x2 )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
213A)
-kx=10x2-100x
or, -k=10x-100
since, the body is displaced by a small distance so we can neglect 10x
therefore,
k=100
now, we know that k=ω2m
therefore,
ω2m=100
or,ω=10√5
now,
T=2π/ω
or,T=π/5√5
B)
actual force=10x2-100x=(16/1000)-4
assumed force=-kx=-4
therefore,
error in calculation=[(16/1000)-4-(-4)/(4-16/1000)]100
=0.4%
Himanshu Giria thanx
466(B) if the amplitude is 4. 0 centimetre, by how much do we error
in assuming that f(x)=- kx at the end point of the motion
213Time period = π/5√5 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 ...