Question

The Function $x=A{\sin }^{2}wt+B{\cos }^{2}wt+C\sin wt\cos wt$ represents SHM for which of the option(s)?

• A. for all value of $A$, $B$ and $C$$(C\ne 0)$
• B. $A=B$, $C=2B$
• C. $A=-B$, $C=2B$
• D. $A=B$, $C=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A for all value of $A$, $B$ and $C$$(C\ne 0)$
B $A=B$, $C=2B$
C $A=-B$, $C=2B$

For representing SHM, the sine and cosine functions should have linear power.
$\therefore x=Asi{n}^2\omega t+Bco{s}^2\omega t+Csin\omega tcos\omega t$ or
$x=\frac{A}{2}\left(2si{n}^2\omega t\right)+\frac{B}{2}\left(2co{s}^2\omega t\right)+\frac{C}{2}\left(2sin\omega tcos\omega t\right)$
or
$x=\frac{A}{2}(1-cos2\omega t)+\frac{B}{2}(1+cos2\omega t)+\frac{C}{2}sin\left(2\omega t\right)$

(a) For $A=0,B=0,x$ $=\frac{C}{2}sin\left(2\omega t\right)$
The equation represents SHM. The option (a) is correct.

(b) If $A=B,C=2B,x=B+Bsin2\omega t$
The equation represents SHM.
The option (b) is correct.

(c) $A=-B,C=2B$,
$\therefore x=Bcos2\omega t+Bsin2\omega t$
or $x=B(cos2\omega t+sin2\omega t)$
Two SHM's are superimposed to give another SHM.
The option (c) is correct.

(d) $A=B,C=0\therefore x=A$
This equation does not represent SHM. Hence options (a), (b) and (c) are correct.