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ONE OR MORE THAN ONE CHOICE CORRECT
A free particle to move along the x-axis has potential energy given by 
where x can be any real number, k being a positive constant, then
(a) at points away from the origin, the particle is in unstable equillibrium.
(b) for any finite non-zero value of x,there is a force directed away from the origin.
(c) if its total mechanical energy is k/2, its maximum kinetic energy is at the origin.
(d) for small displacements from x=0, the motion is shm.
Please please give the explanation for each case. This is a request from my side.
Rates assured.
Comments (7)
Varun, I did all that.
I have used F=-dU/dx. I also used the concept that Mechanical Energy=Kinetic Energy + Potential Energy etc..... etc.....
Can you please give your solution???
And also tell me which of the options are correct, with some explanation for each part.
Arre yaar, option (b) is incorrect according to me. The force is directed towards the origin.
I told you that I did this question and was getting (a), (c), (d) as the answer. But I want someone to confirm the answer because there's a doubt regarding the answers.
Ok now to clearify the options first we have to calculate the force.............
F = -Du/Dx........
F= -2kxe^-x^2......
from here force is always opposite to displace ment..that is if we move to left of origin force is right and for right it will be left..hence option b is incorrect because it give the uniquness of direction..
Du/Dx = 0..then x =0...hence U is function of x and its first derivative is zero at x=0 hene at x=0 particle have either maximum value of P.E or Minimum...
for this differentiate Du/Dx once again w.r.t x..you get positive at x=0..
it means at origin..particle has..minimum potential energy..so KE will be maximum...c is correct.....
potential energy is minimum so it will be stable equilibirium...
so option correct is c....
for shm ...consider the original equation...U will be k(1-1+x^2)....because x is small...
hence U will be kx^2..
so forrce will be -Du/Dx......................................F= -2kx..i.e SHM..so correct optiion is c and d
Ok now to clearify the options first we have to calculate the force.............
F = -Du/Dx........
F= -2kxe^-x^2......
from here force is always opposite to displace ment..that is if we move to left of origin force is right and for right it will be left..hence option b is incorrect because it give the uniquness of direction..
Du/Dx = 0..then x =0...hence U is function of x and its first derivative is zero at x=0 hene at x=0 particle have either maximum value of P.E or Minimum...
for this differentiate Du/Dx once again w.r.t x..you get positive at x=0..
it means at origin..particle has..minimum potential energy..so KE will be maximum...c is correct.....
potential energy is minimum so it will be stable equilibirium...
so option correct is c....
for shm ...consider the original equation...U will be k(1-1+x^2)....because x is small...
hence U will be kx^2..
so forrce will be -Du/Dx......................................F= -2kx..i.e SHM..so correct optiion is c and d
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nice sum
first
solving we get
THE REST CAN BE SOLVED NOW I GUESS