$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.
(N/A) The resultant displacement $x = x_1 + x_2$. At $t = 0$,$x_1 = A_1 \sin(0) = 0$ and $x_2 = A_2 \sin(\frac{\pi}{3}) = A_2 \frac{\sqrt{3}}{2}$. Thus,the displacement is $x = \frac{\sqrt{3}}{2} A_2$.
$(b)$ The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\frac{\pi}{3})}$. Since $\cos(\frac{\pi}{3}) = 0.5$,$A = \sqrt{A_1^2 + A_2^2 + A_1 A_2}$. The maximum speed is $v_{\max} = A \omega = \omega \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(c)$ The maximum acceleration is $a_{\max} = A \omega^2 = \omega^2 \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(b)$ The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\frac{\pi}{3})}$. Since $\cos(\frac{\pi}{3}) = 0.5$,$A = \sqrt{A_1^2 + A_2^2 + A_1 A_2}$. The maximum speed is $v_{\max} = A \omega = \omega \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(c)$ The maximum acceleration is $a_{\max} = A \omega^2 = \omega^2 \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
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