In SHM given by $x = x_m \cos(\omega t + \phi)$, where is the magnitude of the acceleration greatest? at extreme point of $+x_m$ or $-x_m$ 0 between 0 and extreme point
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Step 1: The given equation is $x = x_m \cos(\omega t + \phi)$, where $x_m$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. Show more…
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Assertion The amplitude of a particle executing SHM with a frequency of $60 \mathrm{~Hz}$ is $0.01 \mathrm{~m}$. The maximum value of acceleration of the particle is $\pm 144 \pi^{2} \mathrm{~ms}^{-2}$. Reason Acceleration amplitude $=\omega^{2} A$,where $A$ is displacement amplitude.
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