A particle is vibrating in SHM with an amplitude of 4cm.At what displacement from the equilibrium position will it have mechanical energy which is half potential and half kinetic?
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The formula for KE and PE in SHM is
KE=0.5*m*omega^2*(A^2-x^2) where X is displacement from mean position
PE= 0.5*m*onega^2*x^2.
TE (KE+PE)= 0.5*m*omega^2*A^2.
As PE=KE(both half). From above equations
A^2-x^2=x^2
So. 2x^2=A^2
X=2✓2cm
KE=0.5*m*omega^2*(A^2-x^2) where X is displacement from mean position
PE= 0.5*m*onega^2*x^2.
TE (KE+PE)= 0.5*m*omega^2*A^2.
As PE=KE(both half). From above equations
A^2-x^2=x^2
So. 2x^2=A^2
X=2✓2cm
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