Question

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

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

Answer 1

Answer: (A), (B), (D)

The correct options are
A For any value of $A,B$ and $C$ (except $C=0$)
B If $A=-B,C=2B$ and amplitude $=|B\sqrt{2}|$
D If $A=B,C=2B$ and amplitude $=|B|$

If $A=-B\&C=2B$
$x=B({\cos }^2\omega t-{\sin }^2\omega t)+B\sin 2\omega t$
$\Rightarrow x=B\cos 2\omega t+B\sin 2\omega t$
By multiplying and dividing with $\sqrt{2}$ on both sides ,
$x=\sqrt{2}\times B\times [\frac{1}{\sqrt{2}}\cos 2\omega t+\frac{1}{\sqrt{2}}\sin 2\omega t]$
$\because \sin (A+B)=\sin A\cos B+\cos A\sin B$
we can write the above equation as,
$x=B\sqrt{2}\left[\sin (2\omega t+\frac{\pi }{4})\right]$
This is an expression for particle executing SHM.
Amplitude of $SHM=|B\sqrt{2}|$

If $A=B,C=2B$,
$x=B({\sin }^2\omega t+{\cos }^2\omega t)+B\sin 2\omega t=B+B\sin 2\omega t$
Comparing the above equation with $x=x_0+A^{\prime }\sin {\omega }^{\prime }t$,
We can say that it is also expression for SHM about point $x=B$.
The function oscillates between $x=0\&x=2B$ with an amplitude $B$.

If $A=B\&C=0$, we get $x=B$
This is clearly not an expression for SHM.
Hence, options (b) and (d) are the correct answers.