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 :

Options

A. 5 cm
B. `5sqrt(2)` cm
C. `5sqrt(3)` cm
D. `(10)/(sqrt(2))` cm.

Answer

C

Explanation

`v_("max")=r omega`.
`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 ).