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(C) Given: Amplitude $a = 2 \, cm$,displacement $x = 1 \, cm$.Magnitude of velocity $v = \omega \sqrt{a^2 - x^2} = \omega \sqrt{2^2 - 1^2} = \omega \s

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$\frac{\sqrt{3}}{2\pi}$

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(C) Given: Amplitude $a = 2 \, cm$,displacement $x = 1 \, cm$.Magnitude of velocity $v = \omega \sqrt{a^2 - x^2} = \omega \sqrt{2^2 - 1^2} = \omega \sqrt{3} \, cm/s$.Magnitude of acceleration $A = \omega^2 x = \omega^2 (1) = \omega^2 \, cm/s^2$.According to the problem,the numerical values are equal: $A = v$.$\omega^2 = \omega \sqrt{3} \implies \omega = \sqrt{3} \, rad/s$.Since $\omega = 2\pi f$,we have $f = \frac{\omega}{2\pi} = \frac{\sqrt{3}}{2\pi} \, s^{-1}$.

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