Question

A particle executing SHM has a maximum speed of 30 cm `s^(-1)` and a maximum acceleration of 60 cm `s^(-1)`. The period of oscillation is
A. `pis`
B. `(pi)/(2)s`
C. `2pis`
D. `(pi)/(t)s`

Answer 1

Correct Answer - A
Maximum speed, `v_(max)=omegaA` . .. (i)
Maximum acceleration, `a_(max)=omega^(2)A` . . (ii)
Divide (ii) by (i), we get
`(a_(max))/(v_(max))=(omega^(2)A)/(omegaA)=omega therefore(a_(max))/(v_(max))=(2pi)/(T),T=2pi((v_(max))/(a_(max)))`
here, `v_(max)=30cms^(-1),a_(max)=60cms^(-2)`
`thereforeT=2pi((30cms^(-1))/(60cms^(-2)))=pis`