Given, `m=2kg` , amplitude A=10cm , T=1 s
`(i) omega=(2pi)/(T)=2pis^(-1)=6.28 s^(-1)`
(ii) `v_("max")=Aomega=(10 cm)(2pis^(-1))=0.628 ms^(-1)`
`(iii)a_("max")=Aomega^(2)=(10 cm)(2 pis^(-1))^(2)=4 ms^(-2) " "("take" , pi^(2)=10)`
(vi) `F_("max")=ma_("max")=mAomega^(2)=(2kg)(4 ms^(-2))=8N`
(v) `v=omegasqrt(A^(2)-x^(2))=(6.28 s^(-1))sqrt((10 cm)^(2)-(8 cm)^(2))`
`=(6.28 s^(-1))(6 cm)`
=37.68 cm/s
(vi) Suppose `x=A"cos"omegat`, then the particle will be at the extreme position at time t=0, `v=-Aomega"sin"omegat`
`therefore " At "t=(1)/(12) s, v=-(10 cm)(6.28 s^(-1))"sin"(2pis^(-1).(1)/(12)s)`
`=(6.28 cms^(-1))"sin"(pi)/(6)=-31.4 cms^(-1)`
Negative sign indicates that velocity is directed towards the mean position if the particle starts to the move from the extreme right.
(vii) When time is taken from the mean position, we take
`x=A"sin"omegat`
Suppose, the particle reaches `x=+(A)/(2)` , at time t, then
`(A)/(2)=A"sin"omegatimplies"sin"omegat=(1)/(2)implies omegat=(pi)/(6) s`
or , `t=(pi)/(6 omega)=(pi)/(6xx2pi//T)=(T)/(12)=(1 s)/(12)`
(viii) When time is taken from the extreme position , we take `x=A"cos"omegat`
At t=0, x=A, i.e., the particle is at the extreme right
Suppose at time t, the particle reaches `x=(A)/(2)`, then
`(A)/(2)=A"cos"omegat`
`implies"cos"omegat=(1)/(2)impliesomegat=(pi)/(3)`
or `t=(pi)/(3omega)=(pi)/(3((2pi)/(T)))=(T)/(6)=(1)/(6)s`
