A particle which is attached to a spring oscillates in a horizontal plane with a frequency of `(1)/(pi)Hz` and total energy of 5 J. What is the force constant of the spring, if its maximum speed during oscillation is 40 cm/s ?

Options

A. `150 N//m`
B. `200 N//m`
C. `225 N//m`
D. `250 N//m`

Answer

D

Explanation

`V_("max")=omegaA=2pi n A`
`therefore A= (V_("max"))/(2pi n)=(0.4(m//s))/(2xx pi xx (1)/(pi))`
`=0.2 m = 2xx10^(-1)m`
and `because E= (1)/(2)KA^(2)` for a spring
`therefore K=(2E)/(A^(2))=(2xx5)/((2xx10^(-1))^(2))=(10)/(4xx10^(-2))=250 N//m`