Question

The resultant amplitude due to superposition of two SHMs, $x_1=10sin(\omega t+{30}^{\circ }$)cm and $x_2$= 10 cos ($\omega t+{60}^{\circ }$)cm is

• A. $5cm$
• B. $10cm$
• C. $20cm$
• D. $zero$

Answer 1

The correct option is B $10cm$

$x_1=10sin(wt+{30}^0)$

$x_2=10cos(wt+{60}^0)=-sin(wt-30)=sin(wt+150)$

Thus, we get the two vectors of magnitude 10 inclined at angles ${30}^o$ and ${150}^o$.

The included angle between them is ${120}^o$.
Thus the resultant is $\sqrt{{10}^2+{10}^2+2\left(10\right)\left(10\right)cos120}=10cm$