Q.A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration $\frac{1}{30}$ s after it has crossed the mean position.

[!TLDR] A = 0.04 m, $\omega=2\pi(5)=10\pi$ rad/s. At $t=1/30$ s (from mean position), $\omega t=\pi/3$ rad. $|a|=\omega^2A\sin(\omega …

The needle's amplitude is $A=4$ cm $=0.04$ m and frequency $n=5$ Hz, giving $\omega=2\pi n=10\pi\approx31.416$ rad/s. Since the needle 'crossed the mean position' at t = 0, its motion is $x=A\sin(\omega t)$ and acceleration is $a=-\omega^2x=-\omega^2A\sin(\omega t)$. At $t=\frac{1}{30}$ s, the phase is $\omega t=31.416\times\frac{1}{30}=1.0472$ rad $=60°$, so $\sin(\omega t)=\sin60°=0.8660$. The magnitude of the acceleration …

Take $x=A\sin(\omega t)$ (particle starts at mean position), compute the phase $\omega t$ at the given ti …

Working the phase angle in degrees while leaving $\omega$ in rad/s (a units mismatch), or forgetti …