Question

The displacement of a harmonic oscillator is given by $x=a\sin \omega t+\beta \cos \omega t$. The amplitude of the oscillator is

• A. $a$
• B. $\beta$
• C. $a+\beta$
• D. $\sqrt{a^2+{\beta }^2}$

Answer 1

The correct option is D $\sqrt{a^2+{\beta }^2}$

$x=a\sin \omega t+\beta \cos \omega t$
or, $x=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin \omega t+\frac{\beta }{\sqrt{a^2+b^2}}\cos \omega t)$

Thus, $x=\sqrt{a^2+b^2}\left(sin\right(\omega t+\theta \left)\right)$ (using the property, sin(x+y)=sin x cos y + sin y cos x)

Thus it's of form $x=Asin\omega t$, where $A=\sqrt{a^2+b^2}$