Questions
Consider a body executes periodic motion given by Then we can concluded that
detailed solution
Correct option is A
The given equation 2sin2t+π4 can be compared with y=A sin(ω t+ϕ). This shows that the body executes a simple harmonic motion. Time period can be determined from 'ω' of the given equation. On comparison of 2sin2t+π4 with A sin(ω t+ϕ)., We get ω=2 rad s−1,A=2mWe know that ω=2πt=2πT&ϕ=π4∴2=2πTT=πsHence, We condude that time period T=πs amplitude A=2m and phase ϕ=π4Talk to our academic expert!
Similar Questions
A body executes S.H.M. under the action of a force F1 with a time period 7/6 seconds. If the force is changed to F2 it executes S.H.M. with time period 7/8 seconds. If both the forces F1 and F2 act simultaneously in the same direction on the body, then its time period in seconds is
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