$\text{y}_1=\text{r}\sin\text{wt},\ \text{y}_2=\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)$
From the principle of superposition$\text{y}=\text{y}_1+\text{y}_2$
$=\text{r}\sin\text{wt}+\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)$
$=\text{r}\Big[\sin\text{wt}+\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)\Big]$
$=\text{r}\Bigg[2\sin\bigg\{\frac{\big(\text{wt}+\text{wt}+\frac{\pi}{2}\big)}{2}\bigg\}\cos\bigg\{\frac{\big(\text{wt}-\text{wt}-\frac{\pi}{2}\big)}{2}\bigg\}\Bigg]$
$\Rightarrow\text{y}=2\text{r}\sin\Big(\text{wt}+\frac{\pi}{4}\Big)\cos\Big(\frac{-\pi}{4}\Big)$
Resultant amplitude $=\sqrt{2}\text{r}=4\sqrt{2}\text{mm}$ (because r = 4mm)Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
