Two particles A and B execute simple harmonic motion according to the equations $$y1$$ $$=$$ $$3$$ $$sin$$ ωt and $$y2$$ $$=$$ $$4sin($$ω$$t+$$π$$2)+3sin$$ ωt. The phase difference between them is

Question

Two particles A and B execute simple harmonic motion according to the equations $$y1$$ $$=$$ $$3$$ $$sin$$ ωt and $$y2$$ $$=$$ $$4sin($$ω$$t+$$π$$2)+3sin$$ ωt. The phase difference between them is

Infinity Learn · Accepted Answer

Representing a quantity with phase $$($$ω$$t)$$ along $$X-axis$$ any quantity with a phase $$($$ω$$t+$$ϕ$$)$$ can be represented by a vector making an angle ϕ with $$X-axis$$ in the anticlockwise sense and any quantity with a phase $$($$ω$$t-$$ϕ$$)$$ can be represented by a vector making an angle ϕ with $$X-axis$$ in the clockwise sense.Phase difference ϕ $$=$$ $$tan$$$$-1(43)Alternatively$$, given $$y1$$ $$=$$ $$3$$ $$sin$$ ω$$t-----(i)And$$ $$y2$$ $$=$$ $$4$$ $$sin$$$$($$ω$$t+$$π$$2)+3$$ $$sin$$ ω$$t=$$ $$4$$ $$cos$$ ω$$t+3$$ $$sin$$ ωt $$=$$ $$5$$[$$45cos$$ ω$$t+35sin$$ ωt]$$=5$$[$$sin$$ ϕ $$cos$$ ω$$t+$$$$cos$$ϕ $$sin$$ωt]$$=$$ $$5$$ $$sin$$$$($$ω$$t+$$ϕ$$)-----------------(ii)$$$$cos$$ ϕ $$=$$ $$35$$; $$sin$$ ϕ $$=$$ $$45$$ and $$tan$$ ϕ $$=$$ $$43From$$ (i) and (ii), the result follows.

Two particles A and B execute simple harmonic motion according to the equations $y_{1} = 3 \sin ωt and y_{2} = 4 \sin (ωt + \frac{π}{2}) + 3 \sin ωt$ . The phase difference between them is

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Two particles A and B execute simple harmonic motion according to the equations y1 = 3 sin ωt and y2 = 4sin(ωt+π2)+3sin ωt. The phase difference between them is

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Two particles A and B execute simple harmonic motion according to the equations $y_{1} = 3 \sin ωt and y_{2} = 4 \sin (ωt + \frac{π}{2}) + 3 \sin ωt$ . The phase difference between them is