InterviewSolution
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InterviewSolution
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Two simple harmonic motion are represent by the following equations `y_(1) = 10 sin (pi//4) (12 t + 1)` `y_(2) = 5 (sin 3 theta t + sqrt3 cos 3 theta t)` Here `t` is in seconds. Find out the ratio of their amplitudes.What are the time period of the two motion? |
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Answer» Given equation are `y_(1) = 10 sin (pi//4) (12t + 1)` (i) `y_(2) = 5 (sin 3 pi t + sqrt3 cos 3 pi t)` (ii) We recast these in the form of standard equation of SHM which is `y = A sin (omega t + phi)` (iii) Equation (i) may be written as `y_(1) = 10 sin [(12 pi//4) + (pi//4)]` `= 10 sin (3 pit + pi//4)]` (iv) Comparing Eq (iv) with Eq (iii), we have Amplitude of first SHM `= A_(1) = 10 cm//s and omega_(1) = 3 pi` `:.` Time period of first motion `T_(1) = 2 pi // omega_(1) = 2pi//3pi = (2//3) s` Let us put `5 = A_(2) cos phi` (v) and `5sqrt3 = A_(2) sin phi` (vi) Then `y_(2) = A_(2) cos phi sin (3 pi t + A_(2) sin phi cos 3 pi t)` `= A_(2) sin (3 pi t+ phi)` (vii) Squaring (5) and (6) and adding , we have `A_(2) = sqrt([(5)^(2) + (5sqrt3)^(2)]) = 10 cm` i.e., amplitude of second SHm is `10 cm` and time period of second AHM is `T_(2)` `2 pi// omeag_(2) = 2 pi//3 pi = 2//3 s` Thus, the ratio of amplitudes `A_(1) : A_(2) = 1:1` and perodic times are `T_(1) =T_(2) = (2//3) s` |
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