A simple pendulum starts oscillating simple harmonically from its mean position ( x = 0 ) with amplitude ' a ' and periodic time ' T '. The magnitude of velocity of pendulum at X =

Question

A simple pendulum starts oscillating simple harmonically from its mean position ( x = 0 ) with amplitude ' a ' and periodic time ' T '. The magnitude of velocity of pendulum at X = 2 a ​ is

TestHub · Accepted Answer

The velocity of a particle executing simple harmonic motion at a displacement $x$ from the mean position is given by

$v = ω a^{2} - x^{2}$

The angular frequency is $ω = \frac{2 π}{T}$ .

Substituting $x = \frac{a}{2}$ and $ω = \frac{2 π}{T}$ into the velocity equation:

$v = \frac{2 π}{T} a^{2} - (\frac{a}{2})^{2}$

$v = \frac{2 π}{T} a^{2} - \frac{a ^{2}}{4}$

$v = \frac{2 π}{T} \frac{3 a ^{2}}{4}$

$v = \frac{2 π}{T} (\frac{3 a}{2})$

$v = \frac{3 πa}{T}$

Answer: $\frac{3 πa}{T}$

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