Function x=A sin^2 ω t+B cos^2 ω t+C sin ω t cos ω t represents SHM.

Correct answer: (B) If A=-B, C=2 B , amplitude =|B √2|; (D) If A=B ; C=2 B , amplitude =|B|. For A=-B and C=2 B X=B cos 2 ω t+B sin 2 ω t=√(2 B) sin (2 ω t+(π)/4 ) This is equation of SHM of amplitude √2 B If A=B and C=2 B , then X=B+B sin 2 ω t This is also equation of SHM about the point X=B . Function oscillates between X=0 and X=2 B with amplitude B .

Q99 / 120 · JEE Advanced 2006 · Physics

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 $C=0$ )
Correct answer
B
If $A=-B, C=2 B$, amplitude $=|B \sqrt{2}|$
C
If $A=B ; C=0$
Correct answer
D
If $A=B ; C=2 B$, amplitude $=|B|$

Solution

For $A=-B$ and $C=2 B$ $$ X=B \cos 2 \omega t+B \sin 2 \omega t=\sqrt{2 B} \sin \left(2 \omega t+\frac{\pi}{4}\right) $$ This is equation of SHM of amplitude $\sqrt{2} B$ If $A=B$ and $C=2 B$, then $X=B+B \sin 2 \omega t$ This is also equation of SHM about the point $X=B$. Function oscillates between $X=0$ and $X=2 B$ with amplitude $B$.

Hence, the correct answer is options: 'B', 'D'.

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

Solution

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