If x, v and a denote the displacement,the vel | vedclass

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(B) For a particle executing Simple Harmonic Motion $(SHM)$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{

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$\frac{aT}{x}$

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(B) For a particle executing Simple Harmonic Motion $(SHM)$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$.From the acceleration equation,we have $\frac{a}{x} = -\omega^2$.Since $\omega = \frac{2\pi}{T}$,we have $\omega^2 = \frac{4\pi^2}{T^2}$.Substituting this into the ratio,we get $\frac{a}{x} = -\frac{4\pi^2}{T^2}$.Rearranging this,we get $\frac{aT^2}{x} = -4\pi^2$,which is a constant.However,looking at the options provided,we evaluate $\frac{a}{x} = -\omega^2$. Since $\omega$ and $T$ are constants for a given $SHM$,the ratio $\frac{a}{x}$ is constant. Thus,$\frac{aT}{x}$ is also a constant because $T$ is constant.

If $x, v$ and $a$ denote the displacement,the velocity and the acceleration of a particle executing simple harmonic motion of time period $T$,then,which of the following does not change with time?

Similar Questions

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

If the acceleration of a particle at a distance of $3 \ cm$ from the equilibrium position is $12 \ cm/sec^2$,what is the time period of the simple harmonic motion in seconds?

For a particle executing $SHM$, determine the ratio of the average acceleration of the particle between the extreme position and the equilibrium position with respect to the maximum acceleration.

$Assertion :$ In $SHM$,acceleration is always directed towards the mean position.
$Reason :$ In $SHM$,the body has to stop momentarily at the extreme position and move back to the mean position.

For a particle executing simple harmonic motion,which of the following statements is not correct?

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