The displacement of a particle executing SHM | vedclass

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Question

(C) The displacement of the particle is given by the equation: $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) The displacement of the particle is given by the equation: $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Comparing this with the given equation $y = 0.25 \sin(200t)$:Amplitude $A = 0.25 \ cm$Angular frequency $\omega = 200 \ rad \ s^{-1}$The velocity $v$ of the particle is the time derivative of displacement:$v = \frac{dy}{dt} = \frac{d}{dt} [0.25 \sin(200t)] = 0.25 	imes 200 \cos(200t) = 50 \cos(200t)$The maximum speed $v_{max}$ occurs when $\cos(200t) = 1$:$v_{max} = A \omega = 0.25 	imes 200 = 50 \ cm \ s^{-1}$.

The displacement of a particle executing $SHM$ is given by $y = 0.25 \sin(200t) \ cm$. The maximum speed of the particle is $......... \ cm \ s^{-1}$.

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