The position and velocity of a particle executing simple harmonic motion at t = 0 are given by 3 cm and 8 cm s - 1 respectively. If the angular frequency of the particle is 2 rad s - 1 , then the ampl

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The position and velocity of a particle executing simple harmonic motion at t = 0 are given by 3 cm and 8 cm s - 1 respectively. If the angular frequency of the particle is 2 rad s - 1 , then the amplitude of oscillation (in cm ) is

TestHub · Accepted Answer

Given, the position and velocity of the particle executing $SHM$ .

$y = 3 cm$

$v = 8 cm s^{- 1}$

Angular frequency, $ω = 2 rad s^{- 1}$

The velocity,

$v = ω \sqrt{a^{2} - y^{2}}$

$8 = 2 \sqrt{a^{2} - {(3)}^{2}}$

$4 = \sqrt{a^{2} - {(3)}^{2}}$

$16 = a^{2} - 9$

$a^{2} = 25$

$a = 5 cm$

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