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InterviewSolution
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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. |
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Answer» 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` |
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