Not Just Simple SHM

Electricity and Magnetism Level 4

In the absence of a gravitational field and an external electric field, a thin rod carrying uniform charge − q -q is placed symmetrically along the axis of a thin fixed ring of radius R R . The ring carries a uniformly distributed charge Q Q . The mass of the rod is M M and its length is 2 R 2R . The rod is displaced slightly along the axis of the ring and then released. If the time period T T of small amplitude oscillation is 4 π R ( b a π ϵ M R Q q 4\pi R\sqrt{\frac{(b\sqrt{a}\pi\epsilon MR}{Qq}} , b b and a a being natural numbers, find b + a b+a .


The answer is 4.

Sidharth Soundararajan by Sidharth Soundararajan , 21, India

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1 solution

Rudraksh Shukla
Apr 24, 2016

My expression is coming out to be 4 π R 2 2 ϵ π M R Q q 4\pi{R}\sqrt{\dfrac{2\sqrt{2}\epsilon\pi{MR} }{Qq}} Which means a = 2 a=2 and b = 2 b=2 but the answer is given as a + b a+b i.e. 4 4 rathar than what has been mentioned in the question which is a 2 + b 2 a^2+b^2 which should be the answer i.e. 8 8 . Please check the question.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for pointing that out! I changed it as soon as i could.

Sidharth Soundararajan - 5 years, 1 month ago

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