Simple Harmonic Motion Ques 58
- 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) If $A=-B, C=2 B$, amplitude $=|B \sqrt{2}|$
(c) If $A=B ; C=0$
(d) If $A=B ; C=2 B$, amplitude $=|B|$
$(2006,5 M)]$
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Answer:
Correct Answer: 58.(b,d)
Solution:
- For $A=-B$ and $C=2 B$
$ \begin{aligned} X & =B \cos 2 \omega t+B \sin 2 \omega t \\ & =\sqrt{2} B \sin (2 \omega t+\frac{\pi}{4}) \end{aligned} $
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 $B$.
BETA
