An object executes periodic motion which is given by $$sin4$$$$π$$t−$$cos$$ $$4$$$$π$$t. Then we can say that

Question

An object executes periodic motion which is given by  $$sin4$$$$π$$t−$$cos$$  $$4$$$$π$$t. Then we can  say that

Infinity Learn · Accepted Answer

The function  $$sin4$$$$π$$t−$$cos$$  $$4$$$$π$$t  will represent a periodic motion, if it is identically  repeated after a fixed interval of time. If this is also a simple harmonic motion.  Then we can uniquely write that as   Asinω $$t+$$ϕ ∴$$sin4$$$$π$$t−cas  $$4$$$$π$$t can be reduced as $$followssin4$$$$π$$t−cas  $$4$$$$π$$$$t=212sin4$$$$π$$t−$$12cos$$ $$4$$$$π$$t ∵$$y=A1sin$$ω$$t-A2cos$$ωt& resultant amplitude $$Anet=A12+A22+2A1A2cos$$ϕ here $$A1=A2=1$$ and ϕ$$=90$$ then $$A=12+12+2$$×$$1$$×$$1$$×$$cos90$$ ⇒$$Anet=2$$ (multiplying$$ & dividing by  $$2) $$sin4$$$$π$$t−cas  $$4$$$$π$$$$t=2cos$$$$π$$$$4sin(4$$$$π$$$$t)$$−$$sin$$$$π$$$$4cos$$  (4$$$$π$$$$t)        ∵$$sin$$$$π$$$$4=$$$$cos$$$$π$$$$4=12$$   The above relation can be written $$assin4$$$$π$$t−cas  $$4$$$$π$$$$t=2sin(4$$π  t−π$$4)$$         ∵sinAcosB−$$cosAsinB=$$$$sin$$(A$$−$$B) The above equation can be compared with   $$y=Asin($$ω $$t+$$ϕ$$)$$ This show that the  function  $$sin4$$$$π$$t−$$cos$$  $$4$$$$π$$t is simple harmonic with its resultant as $$2sin(4$$π  t−π$$4)$$ .  Here ω$$=4$$π  Its time period can be found using we know, angular $$frequency=$$ω$$=2$$$$π$$T           .on comparing Here ω$$=4$$π , ω$$=4$$π          $$2$$$$π$$$$T=4$$π          time period $$=T=12s$$ resultant motion of the object is simple harmonic

An object executes periodic motion which is given by sin4πt−cos 4πt. Then we can say that

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