Question

The displacement of a particle executing simple harmonic motin is given by $y-A_0+Asin\omega t+Bcos\omega t$
The amplitude of its oscillation is given by

• A. $A_0+\sqrt{A^2+B^2}$
• B. $\sqrt{A^2+B^2}$
• C. $\sqrt{A_0^2}+(A+B{)}^2$
• D. A + B

Answer 1

The correct option is B $\sqrt{A^2+B^2}$

The general expression for an SHM is
$y=Amp\ast sin(\omega t+\psi )$

$y=A_0+Asin\omega t+Bcos\omega t$
$y-A_0=Asin\omega t+Bcos\omega t$ which implies that the body is performing SHM about $A_0$

Dividing and multiplying RHS by $\sqrt{A^2+B^2}$,

$y-A_0=\sqrt{A^2+B^2}(\frac{A}{\sqrt{A^2+B^2}}sin\omega t+\frac{B}{\sqrt{A^2+B^2}}cos\omega t)$
which implies that the body is performing SHM about $A_0$

We can use the identity
$sin(p+q)=sinpcosq+cospsinq$

Writing $\frac{A}{\sqrt{A^2+B^2}}=cosq$
$\frac{B}{\sqrt{A^2+B^2}}=sinq$ for some angle $q$

$y-A_0=\sqrt{A^2+B^2}sin\left(\omega t+q\right)$

Comparing with the general expression,

amplitude = $\sqrt{A^2+B^2}$