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(B) The general equation for simple harmonic motion is $x(t) = A \cos(\omega t + \phi)$.Given amplitude $A = 0.16 \, m$ and time period $T = 2 \, s$.T

Vedclass · Accepted Answer

$x = -0.16 \cos (\pi t)$

Vedclass · Answer

(B) The general equation for simple harmonic motion is $x(t) = A \cos(\omega t + \phi)$.Given amplitude $A = 0.16 \, m$ and time period $T = 2 \, s$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.At $t = 0$,the displacement $x = -0.16 \, m$.Substituting these values into the equation: $-0.16 = 0.16 \cos(\pi \cdot 0 + \phi) \implies \cos(\phi) = -1 \implies \phi = \pi$.Thus,the displacement equation is $x = 0.16 \cos(\pi t + \pi)$.Using the trigonometric identity $\cos(	heta + \pi) = -\cos(	heta)$,we get $x = -0.16 \cos(\pi t)$.

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