A point moves rectilinearly according to s = 2/9 sin (pi

A point moves rectilinearly according to $s = \dfrac{2}{9}\sin\dfrac{\pi t}{2}+s_{0}$ (cm, s). Find the acceleration $w$ at the end of the first second.

Step 1: Velocity

The chain rule contributes the factor $\dfrac{\pi}{2}$: \[v = \dot{s} = \frac{2}{9}\cdot\frac{\pi}{2}\cos\frac{\pi t}{2} = \frac{\pi}{9}\cos\frac{\pi t}{2}.\]

Step 2: Acceleration

\[ w = \ddot{s} = -\frac{\pi}{9}\cdot\frac{\pi}{2}\sin\frac{\pi t}{2} = -\frac{\pi^{2}}{18}\sin\frac{\pi t}{2}. \]

Step 3: Evaluate at $t = 1$

\[ w(1) = -\frac{\pi^{2}}{18}\sin\frac{\pi}{2} = -\frac{\pi^{2}}{18} \approx -0.5483\ \text{cm/s}^{2}. \]

Step 4: What is happening at $t = 1$

At that instant $v(1) = \dfrac{\pi}{9}\cos\dfrac{\pi}{2} = 0$: the point is momentarily at rest, at its greatest displacement $s = s_{0}+\tfrac29$ from the centre. The acceleration is therefore at its largest magnitude and points back towards the centre — this is the turning point of the oscillation, not a passage through it.

Step 5: Verification

\[w = -\left(\frac{\pi}{2}\right)^{2}\bigl(s-s_{0}\bigr), \text{ the harmonic law with } \omega = \frac{\pi}{2}\]

\[\text{At } t = 1:\ s-s_{0} = \frac29, \text{ so } w = -\frac{\pi^{2}}{4}\cdot\frac29 = -\frac{\pi^{2}}{18} \;\;\checkmark\]

\[\text{Period } T = \frac{2\pi}{\omega} = 4\ \text{s}; \text{ amplitude } \frac29\ \text{cm}; \text{ maximum speed } \frac{\pi}{9} \approx 0.349\ \text{cm/s} \;\;\checkmark\]