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A particle executing simple harmonic motion along y -axis has its motion described by the equation y =A sin(omega t )+ B. The amplitude of the simple harmonic motion is |
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Answer» Solution :Given : `x=A sin (omega t- phi) ""...(1)` `y= B sin omega t""...(2)` In equation (1) use, `sin (A-B)= sin A COS B+ cos A sin B` (1) `rArr x= A sin omega t. + A phi A cos omega t. sin phi x-Asin omega t. cos phi= A cos omega t sin phi` Squaring on both sides we get, `(x-A sin omega t. cos phi)^(2)=A^(2) cos^(2) omega t sin^(2) phi ....(3)` In equation (2) ` sin omega c` can be re- written as, `(Y)/(b)` [ from equation (2)]. Also, use, `cos^(2) omega t=I- sin^(2) omega t` in equation (3) `:. (3)` becomes `(x-A(y)/(B). cos phi)^(2)=A^(2)(1-(y^(2))/(B^(2))) sin ^(2) phi` On EXPANSION. `x^(2)+(A^(2)y^(2))/(B^(2)) cos^(2) phi-(2xAy)/(B) cos phi` `=A^(2) sin^(2) phi-(A^(2)y^(2))/(B^(2)) sin phi....(4)` `x^(2)+(A^(2)y^(2))/(B^(2))( sin^(2) phi+ cos^(2) phi)-(2xyA)/(B) cos phi` `=A^(2) sin^(2) phi "" ( -: byA^(2))` we get, `(x^(2))/(A^(2))+(y^(2))/(B^(2))*1 -(2xy)/(AB) cos phi= sin^(2) phi ""...(5)` Hence proved. Special cases : `varphi=(pi)/(4), "cos" (pi)/(4)=(1)/(SQRT(2)) & "sin" (pi)/(4)=(1)/(sqrt(2))` in equation (5) we get, `(x^(2))/(A^(2))+(y^(2))/(B^(2))-((sqrt(2))XY)/(AB)=(1)/(2)` The above equation is an equation of tilted ellipse. |
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