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The equation of SHM of a particle is given as `2(d^(2)x)/(dt^(2))+32x=0` where x is the displacement from the mean position. The period of its oscillation ( in seconds) is -A. zeroB. `(pi)/(2)`C. `pi`D. `2 pi` |
Answer» Correct Answer - B `(d^(2)x)/(dt^(2))+16x = 0` `therefore" "omega^(2)=16 rArr omega = 4 and T = (2pi)/(omega)=(2pi)/(4)=(pi)/(2)` |
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