Question

Function $x=A{\sin }^2\left(10t\right)+B{\cos }^2\left(10t\right)+Csin\left(10t\right)\cos \left(10t\right)$ represents SHM :

• A. for any value $ofA,B$ and$C$ (except $C=0$)
• B. $ifA=-B;C=2B$, amplitude $=|B\sqrt{2}|$
• C. if A $=B;C=0$
• D. $ifA=B;C=2B$, amplitude $=|B|$

Answer 1

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

$ifA=-B;C=2B$, amplitude $=|B\sqrt{2}|$
$ifA=B;C=2B$, amplitude $=|B|$
for any value $ofA,B$ and$C$ (except $C=0$)

Given : $x=Asi{n}^2\left(10t\right)+Bco{s}^2\left(10t\right)+Csin\left(10t\right)cos\left(10t\right)$

By solving we get, $x=A[\frac{1-cos\left(20t\right)}{2}]$ $+B[\frac{1+cos\left(20t\right)}{2}]+\frac{C}{2}sin\left(20t\right)$

Now $x=[\frac{A}{2}+\frac{B}{2}]$ $+[\frac{A-B}{2}]cos\left(20t\right)+\frac{C}{2}sin\left(20t\right)$ ..........(1)

A particle motion represents $SHM$ whose equation is in the form of $x=Rsin(wt+\varphi )$

$⟹$ for any value of A, B and C (except $C=0$), (1) represents $SHM$.

Example: For $A=B=0,C\ne 0$ $⟹x=\frac{C}{2}sin\left(20t\right)$ which is a $SHM$

But for $A=B,C=0$ $⟹x=B$ which does not represent $SHM$

For $A=-B,C=2B$ $⟹x=Bcos\left(20t\right)+Bsin\left(20t\right)$

$⟹x=B\sqrt{2}[\frac{1}{\sqrt{2}}\times cos\left(20t\right)+\frac{1}{\sqrt{2}}\times sin\left(20t\right)]$

$\therefore x=B\sqrt{2}\left[sin\right({45}^o\left)cos\right(20t)+cos({45}^o\left)sin\right(20t\left)\right]$

$⟹x=\sqrt{2}Bsin(20t+{45}^o)$ represents $SHM$ whose amplitude is $\sqrt{2}B$

For $A=B,C=2B$ $⟹x=B+Bsin\left(20t\right)$

It represents $SHM$ whose amplitude is $B$ and oscillates about a point $x=B$.