Q.
An object executes periodic motion which is given by sin4πt−cos 4πt. Then we can say that
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a
It executes simple harmonic motion with a time period of T = 2s
b
It is periodic but not simple harmonic with a time period of T = 2s
c
It executes simple harmonic motion with a time period of T = ½ s
d
It is periodic but not simple harmonic with a time period of T = ½ s
answer is C.
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Detailed Solution
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
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