Physics - SHM Question with Solution | TestHub

Given equations of SHM are $x_1=a\sin \left(\omega t+\frac{3\pi }{2}\right)$ and $x_2=a\sin (\omega t+\pi )$.

The phase difference between the two SHMs is $\Delta \varphi =\frac{3\pi }{2}-\pi =\frac{\pi }{2}$.

The resultant amplitude $A$ is given by $A=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos (\Delta \varphi )}$.

Substituting $a_1=a$, $a_2=a$, and $\Delta \varphi =\frac{\pi }{2}$:

$A=\sqrt{a^2+a^2+2a^2\cos \left(\frac{\pi }{2}\right)}$

Since $\cos \left(\frac{\pi }{2}\right)=0$, we get:

$A=\sqrt{2a^2}=a\sqrt{2}$

Answer: $a\sqrt{2}$