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A particle executing `SHM` while moving from executy it found at distance `x_(1) x_(2)` and `x_(2)` from comes at the and of three successive second The period of oscilation is where `theta = cos^(-1)((x_1 + x_2)/(2x_(2)))`A. `2 pi//theta`B. `pi//theta`C. `theta`D. `pi//2theta` |
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Answer» Correct Answer - A Displacement time equation of the particle will be `X = A cos omega t` Given that `x_(1) = A cos omega` `and x_(2) = A cos 2omega` `Now (x_(1) + x_(3))/(2x_(2)) = (A (cos theta + cos 3 omega))/(2A cos 2omega)` `= (2A cos 2 omega cos omega)/(2A cos 2 omega)` `= cos omega` `omega = cos^(-1)((x_(1) +x_(2))/(2x_(2))) = (2pi)/(T)` `or T = (2pi)/(theta) where theta = cos^(-1)((x_(1) +x_(2))/(2x_(2)))` |
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