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Question Number 56120 by Tawa1 last updated on 10/Mar/19

Answered by 121194 last updated on 10/Mar/19

without loss of generality, lets takes A paralel to axis x and B dephased θ anti−clockwise A_x =V A_y =0 B_x =Vcos θ B_y =Vsin θ A+B=(A_x +B_x )x^→ +(A_y +B_y )y^→ =(V+Vcos θ)x^→ +Vsin θy^→ ∣A+B∣=(√(V^2 (1+cos θ)^2 +V^2 sin θ)) =V(√((1+cos θ)^2 +sin^2 θ)) =V(√(1+2cos θ+cos^2 θ+sin^2 θ)) =V(√(2(1+cos θ))) cos θ=cos ((θ/2)+(θ/2))=cos^2 (θ/2)−sin^2 (θ/2) cos^2 α+sin^2 α=1⇔sin^2 α=1−cos^2 α cos θ=2cos^2 (θ/2)−1 1+cos θ=2cos^2 (θ/2) ∣A+B∣=V(√(2×2cos^2 (θ/2))) =2V∣cos (θ/2)∣ analogy we have A−B=(A_x −B_x )x^→ +(A_y −B_y )y^→ =V(1−cos θ)x^→ −Vsin θy^→ ∣A−B∣=V(√((1−cos θ)^2 +sin^2 θ)) =V(√(2(1−cos θ))) (1+cos θ)+(1−cos θ)=2 2cos^2 (θ/2)+(1−cos θ)=2 1−cos θ=2(1−cos^2 (θ/2))=2sin^2 (θ/2) ∣A−B∣=2V∣sin (θ/2)∣

$$\mathrm{without}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{generality},\:\mathrm{lets}\:\mathrm{takes}\:\mathrm{A}\:\mathrm{paralel}\:\mathrm{to} \\ $$$$\mathrm{axis}\:{x}\:\mathrm{and}\:\mathrm{B}\:\mathrm{dephased}\:\theta\:\mathrm{anti}−\mathrm{clockwise} \\ $$$${A}_{{x}} ={V} \\ $$$${A}_{{y}} =\mathrm{0} \\ $$$${B}_{{x}} ={V}\mathrm{cos}\:\theta \\ $$$${B}_{{y}} ={V}\mathrm{sin}\:\theta \\ $$$${A}+{B}=\left({A}_{{x}} +{B}_{{x}} \right)\overset{\rightarrow} {{x}}+\left({A}_{{y}} +{B}_{{y}} \right)\overset{\rightarrow} {{y}} \\ $$$$=\left({V}+{V}\mathrm{cos}\:\theta\right)\overset{\rightarrow} {{x}}+{V}\mathrm{sin}\:\theta\overset{\rightarrow} {{y}} \\ $$$$\mid{A}+{B}\mid=\sqrt{{V}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{V}^{\mathrm{2}} \mathrm{sin}\:\theta} \\ $$$$={V}\sqrt{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$={V}\sqrt{\mathrm{1}+\mathrm{2cos}\:\theta+\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$={V}\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\mathrm{cos}\:\theta=\mathrm{cos}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\theta}{\mathrm{2}}\right)=\mathrm{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−\mathrm{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \alpha=\mathrm{1}\Leftrightarrow\mathrm{sin}^{\mathrm{2}} \alpha=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \alpha \\ $$$$\mathrm{cos}\:\theta=\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{cos}\:\theta=\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\mid{A}+{B}\mid={V}\sqrt{\mathrm{2}×\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$=\mathrm{2}{V}\mid\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\mid \\ $$$$\mathrm{analogy}\:\mathrm{we}\:\mathrm{have} \\ $$$${A}−{B}=\left({A}_{{x}} −{B}_{{x}} \right)\overset{\rightarrow} {{x}}+\left({A}_{{y}} −{B}_{{y}} \right)\overset{\rightarrow} {{y}} \\ $$$$={V}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\overset{\rightarrow} {{x}}−{V}\mathrm{sin}\:\theta\overset{\rightarrow} {{y}} \\ $$$$\mid{A}−{B}\mid={V}\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$={V}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$$\left(\mathrm{1}+\mathrm{cos}\:\theta\right)+\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\mathrm{2} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\mathrm{2} \\ $$$$\mathrm{1}−\mathrm{cos}\:\theta=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)=\mathrm{2sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}} \\ $$$$\mid{A}−{B}\mid=\mathrm{2}{V}\mid\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\mid \\ $$

Commented by Tawa1 last updated on 10/Mar/19

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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