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A simple pendulum simple harmonic motion about `x = 0` with an amplitude a and time period `T` speed of the pendulum at `x = a//2` will beA. `pi A sqrt(3) / T`B. `pi A/T`C. `pi A sqrt(3) / 2T`D. `3pi^(2) A/T` |
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Answer» Correct Answer - A `v=omegasqrt(A^(2)-x^(2))=omegasqrt(A^(2)-(A^(2))/(4))` `=omegasqrt((4A^(2)-A^(2))/(2))=(Aomega)/(2)sqrt(3)` `=(sqrt(3))/(2)A(2pi)/(T)=piAsqrt(3)//T` |
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