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A simple pendulum performs simple harmonic motion about x=0x = 0 with an amplitude a'a' and time period T'T'. The speed of the pendulum at x=a2x = \dfrac{a}{2} will be:

হানি নাটস

As simple pendulum performs simple harmonic motion.

velocity,v=ωa2x2\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}

At, x=a2x = \dfrac {a}{2}

v=2πTa2(a2)2=2πT3a22=πa3Tv = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}

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