A particle of mass 0.5g is executing SHM with a time period of 2s and an amplitude of 5 cm. Calculate its (i) maximum velocity, (ii) maximum acceleration and (iii) velocity, acceleration and force acting on the particle when it is at a distance of 4 cm from its position of equilibrium.

Explanation

Amplitude, A = 5cm , time period, T = 2s
`therefore" "omega=(2pi)/T=(2pi)/2=pi"rad"*s^(-1)`
(i) Maximum velocity, `v_(max)=omegaA=pi*5=5pi`
`=5xx3.14=15.7cm*s^(-1)=0.157m*s^(-1)`
(ii) Maximum acceleration, `a_(max)=omega^(2)A=pi^(2)*5`
`=5xx(3.14)^(2)=49.298cm*s^(-2)~~0.493m*s^(-2)`
(iii) When x = 4cm,
velocity, `v=omegasqrt(A^(2)-x^(2))=pisqrt(5^(2)-4^(2))`
`=3.14xx3=9.42cm*s^(-1)`
`=0.094m*s^(-1)`
acceleration, `a=omega^(2)x=(3.14)^(2)xx4`
`=39.438cm*s^(-2)`
`=0.394m*s^(-2)`
and force, F = ma = `0.5xx39.438`
= 19.72 dyn `approx0.197xx10^(-3)N`.