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For a particle performing `SHM`, equation of motion is given as `(d^(2))/(dt^(2)) + 4x = 0`. Find the time periodA. `2pi`B. `(1)/(3)pi`C. `(2)/(3)pi`D. `4 pi` |
Answer» Correct Answer - C Given, `(d^(2)x)/(dt^(2))=-9x` Comparing with `(d^(2)x)/(dt^(2))=-omega^(2)x rArr = 9, omega = 3` `therefore` Time period `= (2pi)/(omega)=(2pi)/(3)=(2)/(3)pi` |
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