Answer
$y= 17 \sin(\omega t + 13^{\circ})$
Work Step by Step
We are asked to find the sum of the two given quantities:
$y_1= 10 \sin \omega t $
$ y_2 =8.0 \sin(\omega t +30^{\circ})$
Now we will set $t=0^{\circ}$ to add them like vectors:
Let $y_h$ be the sum of the horizontal components and $y_v$ be the sum of the vertical components:
$y_h = y_1 \cos t + y_2 \cos t$
$y_h = ( 10 \sin \omega t) \cos t + (8.0 \sin(\omega t +30^{\circ})) \cos t$
$y_h = ( 10 \sin \omega 0^{\circ}) \cos 0^{\circ} + (8.0 \sin(\omega 0^{\circ} +30^{\circ})) \cos 0^{\circ}$
$y_h = 16.9$
$y_v = y_1 \sin t + y_2 \sin t$
$y_v = ( 10 \sin \omega t) \sin t + (8.0 \sin(\omega t +30^{\circ})) \sin t$
$y_v = ( 10 \sin \omega ^{\circ}) \sin ^{\circ} + (8.0 \sin(\omega ^{\circ} +30^{\circ})) \sin 0^{\circ}$
$y_v = 4.0$
Now we can use the Pythagorean theorem to find the magnitud $y_m$:
$y_m=\sqrt{y_h^2+y_v^2}$
$y_m=\sqrt{16.9^2+4.0^2}$
$y_m=17.4$
Now we can use trigonometry to find the angle $\beta$:
$\beta=\arctan (y_v/y_h)$
$\beta=\arctan (4.0/16.9)$
$\beta=13^{\circ}$
Therefore, we have:
$y=y_1=y_2=y_m \sin (\omega t+\beta) = 17 \sin(\omega t + 13^{\circ})$
