A particle of mass 2 kg executing SHM has amplitude 10 cm and time period is 1 s.Find (i) the angular frequency (ii) the maximum speed (ii) the maximum acceleration (iv) the maximum restoring force (v) the speed when the displacement from the mean position is 8 cm (vi) the speed after `(1)/(12)` s the particle was at the extreme position (vii) the time taken by the particle to go directly from its mean position to half the amplitude (viii) the time taken by the particle to go directly from its exterme position to half the amplitude.

Explanation

`m = 2`kg , amplitude `A= 10cm, T = 1s`
(a) `omega = (2pi)/(T) = 2pi s^(1) = 6.28 s^(-1)`
(b) ` v_(max) =A omega ^(2) = (10cm) ( 2pi s^(-1)) = 0.628 ms^(-1)`
(c ) `a_(max) = Aomega^(2) = ( 10 cm ) ( 2pi s^(-1))^(2) = 4ms^(-2) `( take `pi^(2) = 10)`
(d) `F_(max) = ma_(max) mA omega^(2) = ( 2kg) ( 4 ms^(-2)) = 8N`
(e) `v=omegasqrt(A^(2) - x^(2)) = ( 6.28 s^(-1)) sqrt(( 10cm^(2)) - (8 cm)^(2))`
`= ( 6.28 s^(-1))( 6 cm)=37.68 cm//s`
(f) Suppose `x=A cos omegat` , then the particle will be at the extreme position at time `t=0`
`v=- Aomega sin omegat `
`:. ` At ` t= (1)/(12) s , v = - (10 cm ) ( 6.28 s^(-1)) sin ( 2pis^(-1)(1)/(12)s)`
`= - ( 62.8 cm s^(-1)) sin. (pi)/(6) = - 31.4 cms^(-1)`
Negative sign indicated that velocity is directed towards the mean position if the paritcle starts to move from the right extreme.
(g) When time is taken from the mean position , we take
`x = A sin omegat t`
Suppose the particle reaches `x= + (A)/(2)`, at time t , then
`(A)/(2) = A sin omegat implies sin omegat = (1)/(2) implies omegat = (pi)/(6)`
or , ` t= (pi)/(6omega)= (pi)/( 6 xx 2 pi//T) = (T)/(12) = (1s)/( 12) = (1)/(12) s `
(h) When time is taken from the extreme position, we take ` x= A cos omegat `
At `t = 0 , x= A `, i.e., the paritcle is at the right extreme
Suppose at time t, the particle reaches `x = (A)/(2)` then
`(A)/(2) = A cos omegat implies cos omegat =(1)/(2) implies omegat =(pi)/(3)`
or `, t= ( pi)/(3omega) =(pi)/(3((2pi)/T))=(T)/(6)`