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)

If $A=B,C=2B$ and amplitude $=|B|$

Given,
$x=A{\sin }^2\omega t+B{\cos }^2\omega t+C\sin \omega t\cos \omega t$

Re writing the above equation we get,

$x=(\frac{A}{2}+\frac{B}{2})-(\frac{A}{2}-\frac{B}{2})\cos 2\omega t+\frac{C}{2}\sin 2\omega t......\left(1\right)$

If $A=-B\&C=2B$ , then from $\left(1\right)$ we get that ,

$x=B\cos 2\omega t+B\sin 2\omega t$

By multiplying and dividing with $\sqrt{2}$ on RHS,

$x=\sqrt{2}\times B\times [\frac{1}{\sqrt{2}}\cos 2\omega t+\frac{1}{\sqrt{2}}\sin 2\omega t]$

$\because \sin (A+B)=\sin A\cos B+\cos A\sin B$

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 $A=B,C=2B$, from $\left(1\right)$ we can write that,

$x=B+B\sin 2\omega t$

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.

If $A,B\&C$ can take any values except $C=0$ then, we can say from $\left(1\right)$ that it is an expression of SHM of a particle which can oscillate about any point that depends upon the values of $A$ and $B$.

Hence, options (a) , (b) and (d) are the correct answers.