InterviewSolution
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InterviewSolution
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The equation of the resulting oscillation obtained by the summation at two mutally perpendicular oscillation with the same frequency `f_(1) = f_(2) = 5 Hz` and same initial phase `delta_(1) = delta_(2) = 60^(@) ` is (Given their amplitude are `A_(1) = 0.1m and A_(2) = 0.05m`A. `0.15sin(10pi t +(pi)/(6))`B. `0.05sin(10pi t +(2pi)/(3))`C. `0.112 sin(10pi t +(pi)/(3))`D. `0.313 sin(10pi t +(pi)/(2))` |
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Answer» Correct Answer - C When `x = A_(1) sin (omega t + phi_(1))` and `y = A_(2) sin (omegat + phi_(2))` then `(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) cos delta = sin ^(2) delta` Here phase i.e. `delta = 0` `:.(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) = 0` `rArr ((x)/(A_(1)) - (y)/(A_(2)))^(2) = 0 rArr y = (A_(2))/(A_(1)) x` REsulting amplitude is `A = sqrt(A_(1)^(2) + A_(2)^(2)) = sqrt((0.1)^(2) + (0.05)^(2)) = 0.112m` Thus the equation of resulting `SHM` ` r = 0.112 sin (10pi t + (pi)/(3))` |
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