Question

Function $x=A{\sin }^2\omega t+B{\cos }^2\omega t+C\sin \omega t\cos \omega t$ represents SHM

• A. For any value of A ,B and C (except C=0)
• B. $A=-B,C=2B,\text{amplitude}=|B\sqrt{2}|$
• C. $A=B;C=0$
• D. $A=B;C=2B,\text{amplitude}=|B|$

Answer 1

The correct option is D $A=B;C=2B,\text{amplitude}=|B|$

$\begin{array}{c}x=A{\sin }^2\omega t+B{\cos }^2\omega t+C\sin \omega \cos \omega t\\ =\frac{A\left(1-\cos 2\omega t\right)}{2}+\frac{B\left(\cos 2\omega t-1\right)}{2}+\frac{C}{2}\sin 2\omega t\\ =\left(\frac{A}{2}-\frac{B}{2}\right)+\left(\frac{B}{2}-\frac{A}{2}\right)\cos \omega t+\frac{c}{2}\sin \omega t\\ ifA=B\&C=2B\\ x=B+B\sin 2\omega t\\ \therefore Amplitude=B\end{array}$

$\therefore$ Option $D$ is correct.