Question

A particle is subjected to two S H Ms $x_1=A_1$sin wt and $x_2=A_2$ sin $(wt+\pi /4).$ The resultant S H M will have an amplitude of

• A. $\frac{A_1+A_2}{2}$
• B. $\sqrt{A_1^2+A_2^2}$
• C. $\sqrt{A_1^2+A_2^2+\sqrt{2A_1{A}_2}}$
• D. $A_1{A}_2$

Answer 1

The correct option is B $\sqrt{A_1^2+A_2^2}$

Let equation of resultant S H M

$\Rightarrow x=x_1+x_2=A_{1}sin\omega t+A_{2}sin(\omega t+\frac{\pi }{4})$

$\Rightarrow x=\sqrt{A_1^2+A_2^2}(\frac{A_1}{\sqrt{A_1^2+A_2^2}}sin\omega t+\frac{A_2}{\sqrt{A_1^2+A_2^2}}sin(\omega t+\frac{\pi }{4}\left)\right)$

$\Rightarrow x=\sqrt{A_1^2+A_2^2}sin(\omega t+\varphi )$

the amplitude of the SHM of x is $\sqrt{A_1^2+A_2^2}$

Hence the correct option is B