An object executes periodic motion which is given by  $\sin 4\pi t-cos4\pi t.$ Then we can  say that

detailed solution

Correct option is C

The function  sin4πt−cos  4πt  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 ∴sin4πt−cas  4πt can be reduced as followssin4πt−cas  4πt=212sin4πt−12cos 4πt ∵y=A1sinωt-A2cosωt& resultant amplitude Anet=A12+A22+2A1A2cosϕ here A1=A2=1 and ϕ=90 then A=12+12+2×1×1×cos90 ⇒Anet=2 (multiplying & dividing by  2) sin4πt−cas  4πt=2cosπ4sin(4πt)−sinπ4cos  (4πt)        ∵sinπ4=cosπ4=12   The above relation can be written assin4πt−cas  4πt=2sin(4π  t−π4)         ∵sinAcosB−cosAsinB=sin(A−B) The above equation can be compared with   y=Asin(ω t+ϕ) This show that the  function  sin4πt−cos  4πt is simple harmonic with its resultant as 2sin(4π  t−π4) .  Here ω=4π  Its time period can be found using we know, angular frequency=ω=2πT           .on comparing Here ω=4π , ω=4π          2πT=4π          time period =T=12s resultant motion of the object is simple harmonic