Question

A particle is subjected to two SHMs $x_1=A_{1}sin\omega t$ and $x_2=A_2(\omega t+\frac{\pi }{4})$. The resultant SHM will have an amplitude of

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$A=\sqrt{A_1^2+A_2^2+2A_1{A}_{2}cos\theta }=\sqrt{A_2^1+A_2^2+2A_1{A}_{2}cos\frac{\pi }{4}}=\sqrt{A_1^2+A_2^2+\sqrt{2}{A}_1{A}_2}$