In a simple harmonic motion,let f be the acce | vedclass

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(A) In simple harmonic motion,the acceleration $f$ is given by $f = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.T

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(A) In simple harmonic motion,the acceleration $f$ is given by $f = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.Taking the magnitude,we have $|f| = \omega^2 |x|$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Therefore,the product $|fT|$ is:$|fT| = |f| \cdot T = (\omega^2 |x|) \cdot \left(\frac{2\pi}{\omega}\right) = 2\pi\omega |x|$.Since $2\pi\omega$ is a constant,the relationship $|fT| = (2\pi\omega) |x|$ represents a linear equation of the form $y = mx$,where $y = |fT|$ and $x$ is the displacement.Thus,the graph of $|fT|$ versus $x$ is a straight line passing through the origin.

In a simple harmonic motion,let $f$ be the acceleration and $T$ be the time period. If $x$ denotes the displacement,then the $|fT|$ vs. $x$ graph will look like,

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