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Question-58001

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Question Number 58001 by rahul 19 last updated on 16/Apr/19
Commented by rahul 19 last updated on 16/Apr/19
I hope Q. is clear. Ans→ t=(1/(24))s ,(5/(24))s,(7/(24))s,((11)/(24))s,.....
$${I}\:{hope}\:{Q}.\:{is}\:{clear}. \\ $$$${Ans}\rightarrow\:{t}=\frac{\mathrm{1}}{\mathrm{24}}{s}\:,\frac{\mathrm{5}}{\mathrm{24}}{s},\frac{\mathrm{7}}{\mathrm{24}}{s},\frac{\mathrm{11}}{\mathrm{24}}{s},….. \\ $$
Answered by mr W last updated on 16/Apr/19
T=((2π)/ω)=((2π)/(4π))=(1/2) sec extreme position at t=0: θ_1 =(π/2) possibility 1: currect position: sin θ_2 =((√3)/2)⇒θ_2 =(π/3) ⇒t=((θ_1 −θ_2 )/(2π))×T=(((π/2)−(π/3))/(2π))×(1/2)=(1/(24)) sec or generally t=nT±(1/(24)) with n∈N≥0 possibility 2: currect position: sin θ_2 =−((√3)/2)⇒θ_2 =−(π/3) ⇒t=((θ_1 −θ_2 )/(2π))×T=(((π/2)+(π/3))/(2π))×(1/2)=(5/(24)) sec or generally t=nT±(5/(24)) with n∈N≥0
$${T}=\frac{\mathrm{2}\pi}{\omega}=\frac{\mathrm{2}\pi}{\mathrm{4}\pi}=\frac{\mathrm{1}}{\mathrm{2}}\:{sec} \\ $$$${extreme}\:{position}\:{at}\:{t}=\mathrm{0}:\:\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$$${possibility}\:\mathrm{1}: \\ $$$${currect}\:{position}:\:\mathrm{sin}\:\theta_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\theta_{\mathrm{2}} =\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{t}=\frac{\theta_{\mathrm{1}} −\theta_{\mathrm{2}} }{\mathrm{2}\pi}×{T}=\frac{\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}}{\mathrm{2}\pi}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{24}}\:{sec} \\ $$$${or}\:{generally}\:{t}={nT}\pm\frac{\mathrm{1}}{\mathrm{24}}\:{with}\:{n}\in{N}\geqslant\mathrm{0} \\ $$$$ \\ $$$${possibility}\:\mathrm{2}: \\ $$$${currect}\:{position}:\:\mathrm{sin}\:\theta_{\mathrm{2}} =−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\theta_{\mathrm{2}} =−\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{t}=\frac{\theta_{\mathrm{1}} −\theta_{\mathrm{2}} }{\mathrm{2}\pi}×{T}=\frac{\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}}{\mathrm{2}\pi}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{24}}\:{sec} \\ $$$${or}\:{generally}\:{t}={nT}\pm\frac{\mathrm{5}}{\mathrm{24}}\:{with}\:{n}\in{N}\geqslant\mathrm{0} \\ $$
Commented by rahul 19 last updated on 16/Apr/19
why θ_(1 ) cannot be =− 90°/270° ? Doing the above procedure,the time you will get will come out to be −ve. What does this mean?
$${why}\:\theta_{\mathrm{1}\:} {cannot}\:{be}\:=−\:\mathrm{90}°/\mathrm{270}°\:? \\ $$$${Doing}\:{the}\:{above}\:{procedure},{the}\:{time} \\ $$$${you}\:{will}\:{get}\:{will}\:{come}\:{out}\:{to}\:{be}\:−{ve}. \\ $$$${What}\:{does}\:{this}\:{mean}? \\ $$
Commented by mr W last updated on 16/Apr/19
Commented by mr W last updated on 16/Apr/19
θ_1 can be taken as any value. we are only interested at θ_1 −θ_2 . you understand what is meant with t=nT±(1/(24)) and t=nT±(5/(24)). in each periode T there arefour instants at which the position is ((√3)/2) apart from the mean position as shown in the diagram. t is certainly >0. naturally i can write t=nT+(1/(24)) t=nT+(5/(24)) t=nT+(7/(24)) t=nT+((11)/(24))
$$\theta_{\mathrm{1}} \:{can}\:{be}\:{taken}\:{as}\:{any}\:{value}.\:{we}\:{are}\:{only} \\ $$$${interested}\:{at}\:\theta_{\mathrm{1}} −\theta_{\mathrm{2}} . \\ $$$${you}\:{understand}\:{what}\:{is}\:{meant}\:{with} \\ $$$${t}={nT}\pm\frac{\mathrm{1}}{\mathrm{24}}\:{and}\:{t}={nT}\pm\frac{\mathrm{5}}{\mathrm{24}}.\:{in}\:{each} \\ $$$${periode}\:{T}\:{there}\:{arefour}\:{instants}\:{at}\:{which} \\ $$$${the}\:{position}\:{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{apart}\:{from}\:{the}\:{mean} \\ $$$${position}\:{as}\:{shown}\:{in}\:{the}\:{diagram}. \\ $$$${t}\:{is}\:{certainly}\:>\mathrm{0}.\:{naturally}\:{i}\:{can}\:{write} \\ $$$${t}={nT}+\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${t}={nT}+\frac{\mathrm{5}}{\mathrm{24}} \\ $$$${t}={nT}+\frac{\mathrm{7}}{\mathrm{24}} \\ $$$${t}={nT}+\frac{\mathrm{11}}{\mathrm{24}} \\ $$
Commented by rahul 19 last updated on 16/Apr/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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