Physics - SHM Question with Solution | TestHub

${\mathrm{x}}_1=5\sin \mathrm{\omega t}$

${\mathrm{x}}_2=5\sin \left(\mathrm{\omega t}+53°\right)$

${\mathrm{x}}_3=-10\mathrm{cos\omega t}$

we can write ${\mathrm{x}}_3=10\sin \left(\mathrm{\omega t}-90°\right)$ Finding the resultant amplitude by vector notation.

Resultant Amplitude $\left|R\right|=\sqrt{8^2+6^2}=10$.