A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:

\frac{π a}{T}

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\frac{{3 π}^{2} a}{T}

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\frac{π a \sqrt{3}}{T}

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\frac{π a \sqrt{3}}{2 T}

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Solution

The correct option is C $\frac{π a \sqrt{3}}{T}$
$v = ω \sqrt{a^{2} - x^{2}}$
$a =$ amplitude
$x =$ position
$v = \frac{2 π}{T} \sqrt{A^{2} - {(\frac{A}{2})}^{2}}$
$= \frac{2 π}{T} \sqrt{\frac{3 A^{2}}{4}}$
$V > \frac{π A}{T} \sqrt{3}$
Option C.

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A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:

The correct option is C πa√3Tv=ω√a2−x2a= amplitudex= positionv=2πT√A2−(A2)2=2πT√3A24V>πAT√3Option C.

The correct option is C $\frac{π a \sqrt{3}}{T}$
$v = ω \sqrt{a^{2} - x^{2}}$
$a =$ amplitude
$x =$ position
$v = \frac{2 π}{T} \sqrt{A^{2} - {(\frac{A}{2})}^{2}}$
$= \frac{2 π}{T} \sqrt{\frac{3 A^{2}}{4}}$
$V > \frac{π A}{T} \sqrt{3}$
Option C.