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A particle executes simple harmonic motion of amplitude A (i) At what distance from the extreme positino is its kinetic energy equal to half its potential energy ? (ii) At what position from the extreme position is its speed half the maximum speed. |
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Answer» If y is th edisplacement of particle from the mean position, then P.E. of particle , `U=(1)/(2)ky^(2)` and K.E. of particle, `K=(1)/(2)k (A^(2)-y^(2))` where A is the amplitude of oscillation and k is a force constant. As per question, `K=(U)/(2), so` `(1)/(2)k(A^(2)-y^(2))=(1)/(2)((1)/(2)ky^(2))` or `A^(2)-y^(2)=(y^(2))/(2) or 2A^(2)-2y^(2)=y^(2)` or `y=ssqrt((2)/(3))A` ` :.` Distnace of location from extreme position `=A-sqrt((2)/(3))A=A(sqrt(3)-sqrt(2))//sqrt(3)` `=A(1.732-1514)//1.732=0.83A` `=0.183` times the amplitude (ii) `v_(max)=Aomega` and speed at displacement from mean position is `v=omegasqrt(A^(2)-y^(2))` As per question, `v=v_(max)//2` As `omega sqrt(A^(2)-y^(2))=Aomega//2 or A^(2)-y^(2)=A^(2)//4` or `y^(2)=A^(2)-A^(2)=3A^(2)//4` or ` y =sqrt(3)A//2` `:.` Distance from the extreme position is `=A-y=A-(sqrt(3)A)/(2)=A[(2-sqrt(3))/(2)]` `=A[(2-1.732)/(2)]=0.134A` `=0.134` times the amplitude. |
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