A particle is performing U.C.M. along the cirqumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in `m//s^(2)` is

Options

A. `2 pi^(2)`
B. ` 8 pi^(2)`
C. `pi^(2)`
D. `4pi^(2)`

Answer

D

Explanation

(d) Given, diameter of circle ,d=50cm
` " " =50xx10^(-2)m`
and frequency ,f= 2Hz
The acceleration of particle in a uniform circular motion can be given as
` " "a = omega ^(2)x `
where, `omega =` angular frequency = ` 2pi f `
` " " x =` distance from centre ` =(d)/(2)`
`rArr " "a= 4pi ^(2)f^(2) xx(d)/(2) " "...(i)`
Substituting given values in Eq. (i) we get
` " "a= 4pi ^(2)xx 4xx(50 xx10^(-2) )/( 2) =4pi ^(2) `