A particle of mass m is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm s^(-1). Its velocity will be 50 cm s^(-1) at a distance :

A particle of mass m is executing S.H.M. about a point with amplitude

1.	A particle of mass m is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm s^(-1). Its velocity will be 50 cm s^(-1) at a distance :
Answer» 5 CM `5sqrt(2)` cm `5sqrt(3)` cm `(10)/(SQRT(2))` cm. Solution :`v_("max")=ROMEGA`. `100=10.oemgaimpliesomega=10" rad"//"s"`. Now `v^(2)=OMEGA^(2)(r^(2)-y^(2))`. `:.""2500=100(100-y^(2))` `y^(2)=100-25=75` `y=5sqrt(3)` cm. Aliter. We can use the formula that at `y=(sqrt(3))/(2)r` `v=(v_("max"))/(2)` `:.""y=(sqrt(3))/(2)xx10=5sqrt(3)" cm"`. So the correct choice is ( c ).

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A particle of mass m is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm s^(-1). Its velocity will be 50 cm s^(-1) at a distance :

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