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A particle executes SHM with an amplitude 4 cm. At what displacement from the mean position its energy is half kinetic and half potential

Asked by nipunverma59 8th October 2018, 4:39 PM
Answered by Expert
Equation for displacement x of SHM :-   x = A×sin(ωt) .......................... (1)
Equation for velocity v of SHM :-            v = Aω×cos(ωt) .......................... (2)
Maximum kinetic energy :-  begin mathsize 12px style K subscript m a x end subscript space equals space 1 half m v subscript m a x end subscript superscript 2 space equals space 1 half m cross times A squared cross times omega squared space....................... left parenthesis 3 right parenthesis end style
when Kinetic energy is maximum potential energy is zero.
when kinetic energy is half of its maximum, potential energy will become other half and both are equal.
velocity of the partile v in SHM, when its kinetic energy is hallf of maximum is obtained as follows
begin mathsize 12px style 1 half m space v squared space equals space 1 half cross times 1 half m cross times A squared cross times omega squared space space space o r space space v space equals space fraction numerator 1 over denominator square root of 2 end fraction A omega end style  ...........................................(4)
if we substitute the value of v from eqn.(4) in eqn.(2), we get the argument of cosine function i.e., ωt = π/4 .
If the same argument ωt = π/4 is applied in eqn.(1) we get x = A/√2.
Hence in SHM, when the particle is at a distance A/√2 from its mean position, its kinetic energy is equal to its potetial energy
Answered by Expert 8th October 2018, 6:42 PM
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