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Question Number 133590 by physicstutes last updated on 23/Feb/21
A particle performs simple harmonic motion between two points A and B which are 10 m apart on a horizontal straight line. When the particle is 3 m away from the centre, O, of the line AB, its speed is 8 ms^(−1) . Find the least time required for the particle to move from B to the midpoint of OA.
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{performs}\:\mathrm{simple}\:\mathrm{harmonic}\:\mathrm{motion}\:\mathrm{between}\:\mathrm{two}\:\mathrm{points} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{which}\:\mathrm{are}\:\mathrm{10}\:\mathrm{m}\:\mathrm{apart}\:\mathrm{on}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{straight}\:\mathrm{line}. \\ $$$$\mathrm{When}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{3}\:\mathrm{m}\:\mathrm{away}\:\mathrm{from}\:\mathrm{the}\:\mathrm{centre},\:\mathrm{O},\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{AB}, \\ $$$$\mathrm{its}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{8}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{Find}\:\mathrm{the}\:\mathrm{least}\:\mathrm{time}\:\mathrm{required}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{to}\:\mathrm{move}\:\mathrm{from}\:\mathrm{B}\:\mathrm{to}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{OA}. \\ $$
Answered by mr W last updated on 23/Feb/21
v=V_0 cos ωt with V_0 =speed at O s=S_1 sin ωt with S_1 =OB=5 m T=period=((2π)/ω) at t=t_1 : s=3m, v=8 m/s 8=V_0 cos ωt_1 3=5 sin ωt_1 ⇒((8/V_0 ))^2 +((3/5))^2 =1 ⇒V_0 =10 m/s on the other side: V_0 =ωS_1 ⇒ω=(V_0 /S_1 )=((10)/8)=(5/4) 1/s ⇒T=((2π)/ω)=((8π)/5) s least time from B to O =(T/4)=((2π)/5) at midpoint OA=C: s=((OA)/2)=(5/2) (5/2)=5 sin ωt_2 ⇒ωt_2 =sin^(−1) (1/2)=(π/6) ⇒t_2 =((4π)/(5×6))=((2π)/(15)) least time from B to C: (T/4)+t_2 =((2π)/5)+((2π)/(15))=((8π)/(15))≈1.675 s
$${v}={V}_{\mathrm{0}} \mathrm{cos}\:\omega{t}\:{with}\:{V}_{\mathrm{0}} ={speed}\:{at}\:{O} \\ $$$${s}={S}_{\mathrm{1}} \mathrm{sin}\:\omega{t}\:{with}\:{S}_{\mathrm{1}} ={OB}=\mathrm{5}\:{m} \\ $$$${T}={period}=\frac{\mathrm{2}\pi}{\omega} \\ $$$${at}\:{t}={t}_{\mathrm{1}} :\:{s}=\mathrm{3}{m},\:{v}=\mathrm{8}\:{m}/{s} \\ $$$$\mathrm{8}={V}_{\mathrm{0}} \mathrm{cos}\:\omega{t}_{\mathrm{1}} \\ $$$$\mathrm{3}=\mathrm{5}\:\mathrm{sin}\:\omega{t}_{\mathrm{1}} \\ $$$$\Rightarrow\left(\frac{\mathrm{8}}{{V}_{\mathrm{0}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{V}_{\mathrm{0}} =\mathrm{10}\:{m}/{s} \\ $$$${on}\:{the}\:{other}\:{side}:\:{V}_{\mathrm{0}} =\omega{S}_{\mathrm{1}} \\ $$$$\Rightarrow\omega=\frac{{V}_{\mathrm{0}} }{{S}_{\mathrm{1}} }=\frac{\mathrm{10}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{4}}\:\mathrm{1}/{s} \\ $$$$\Rightarrow{T}=\frac{\mathrm{2}\pi}{\omega}=\frac{\mathrm{8}\pi}{\mathrm{5}}\:{s} \\ $$$${least}\:{time}\:{from}\:{B}\:{to}\:{O}\:=\frac{{T}}{\mathrm{4}}=\frac{\mathrm{2}\pi}{\mathrm{5}} \\ $$$${at}\:{midpoint}\:{OA}={C}:\:{s}=\frac{{OA}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{5}\:\mathrm{sin}\:\omega{t}_{\mathrm{2}} \\ $$$$\Rightarrow\omega{t}_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\frac{\mathrm{4}\pi}{\mathrm{5}×\mathrm{6}}=\frac{\mathrm{2}\pi}{\mathrm{15}} \\ $$$${least}\:{time}\:{from}\:{B}\:{to}\:{C}:\: \\ $$$$\frac{{T}}{\mathrm{4}}+{t}_{\mathrm{2}} =\frac{\mathrm{2}\pi}{\mathrm{5}}+\frac{\mathrm{2}\pi}{\mathrm{15}}=\frac{\mathrm{8}\pi}{\mathrm{15}}\approx\mathrm{1}.\mathrm{675}\:{s} \\ $$
Commented by physicstutes last updated on 23/Feb/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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