Two particles A and B execute simple harmonic motion according to the equations $$)$$ $$y1=3sin$$⁡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 $$)$$ $$y1=3sin$$⁡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 $$($$ωt−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 $$y1=3sin$$⁡ωt.     ......(i) and  $$y2=4sin$$⁡ω$$t+$$π$$2+3sin$$⁡ω$$t=4cos$$⁡ω$$t+3sin$$⁡ω$$t=545cos$$⁡ω$$t+35sin$$⁡ω$$t=5$$[$$sin$$⁡ϕ$$cos$$⁡ω$$t+$$$$cos$$⁡ϕ$$sin$$⁡ωt]$$=5sin$$⁡$$($$ω$$t+$$ϕ$$)$$      .........(ii)From$$ $$(i) and (ii), the result follows.$$cos$$⁡ϕ$$=35$$;$$sin$$⁡ϕ$$=45$$ and $$tan$$⁡ϕ$$=43$$

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

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Two particles A and B execute simple harmonic motion according to the equations ) y1=3sin⁡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 ) $y_{1} = 3 \sin ω t$ and $y_{2} = 4 \sin (ω t + \frac{π}{2}) + 3 \sin ω t$ The phase difference between them is