Question

A particle is executing SHM of amplitude A. What fraction of the total energy is kinetic when the displacement is a quarter of the amplitude

Answer 1

Answer:

A particle executing linear SHM with amplitude "a". What fraction of the total energy is kinetic when displacement is half the amplitude?

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As per the question, assuming the particle executes simple harmonic motion with amplitude ‘a’, the equation of motion of the particle is given by:

x=acos(ωt+ϕ)

Hence, the velocity of the particle at any instant ‘t’ is given by

v=dxdt=−aωsin(ωt+ϕ)

So, total energy of the particle at any instant is given by sum of the kinetic and potential energies,

E=KE+PE

=12mv2+12mω2x2

=12m[−aωsin(ωt+ϕ)]2+12mω2[acos(ωt+ϕ)]2

=12mω2a2[sin2(ωt+ϕ)+cos2(ωt+ϕ)]

=12mω2a2

Required fraction of total energy when displacement x=a2 is given by

KEE=12mv212mω2a2

=(−aωsin(ωt+ϕ))2ω2a2

=(−aω)2(1−cos2(ωt+ϕ))ω2a2

=(a2−a2cos2(ωt+ϕ))a2

=(a2−x2)a2

And given that x=a2 ,

KEE=3a24a2=34=75 %

The answer is 75 % of the total energy is in the form of kinetic energy when x=a2.

Answer 2

Answer:

The fraction of Kinetic energy to the total energy of a particle executing SHM when the displacement is a quarter of the amplitude is $93%$%

Explanation:

• For a particle executing simple harmonic motion the kinetic energy of that particle is

               $KE=\frac{1}{2}mw^{2}(A^{2}-x^{2} )$

• The potential energy is given as

            $PE=\frac{1}{2}mw^{2}x^{2}$

• The total energy

           $E=PE+KE=KE=\frac{1}{2}mw^{2}A^{2}$

where

$m$-the mass of the particle

ω- the angular frequency

$A$- amplitude of the oscillation

$x$- displacement

From the question, it is given that

The displacement is a quarter of the amplitude $x=A/4$

substitute this in kinetic energy

$KE=\frac{1}{2}mw^{2}(A^{2}-(A/4)^{2} )\\ KE =\frac{1}{2} mw^{2} A^{2} (15/16)$

the fraction of the kinetic energy in total energy is $KE/E=15/16$ ≈93%