Question

The displacement $y$ of a particle varies with time $t$ according to the relation $y=a\sin \omega t+b\cos \omega t$. Which of the following statement(s) is/are correct regarding the motion?

• A. The motion is SHM.
• B. The motion is SHM with amplitude $a+b$.
• C. The motion is SHM with amplitude $a^2+b^2$.
• D. The motion is SHM with amplitude $\sqrt{a^2+b^2}$.

Answer 1

Answer: (A), (D)

The correct options are
A The motion is SHM.
D The motion is SHM with amplitude $\sqrt{a^2+b^2}$.

From the data given in the question, the displacement
$y=a\sin \omega t+b\cos \omega t....\left(1\right)$
Let $a=A\cos \varphi$ and $b=A\sin \varphi$
Now, $a^2+b^2=A^2{\cos }^2\varphi +A^2{\sin }^2\varphi =A^2$
$\Rightarrow A=\sqrt{a^2+b^2}$
From $\left(1\right)$, we can say that
$y=A\cos \varphi .\sin \omega t+A\sin \varphi .\cos \omega t=A\sin (\omega t+\varphi )$

Double differentiating with respect to time on both sides, we get
$\frac{dy}{dt}=A\omega \cos (\omega t+\varphi )$
$\frac{d^{2}y}{d{t}^2}=-A{\omega }^2\sin (\omega t+\varphi )=-{\omega }^{2}y$
$\Rightarrow \frac{d^{2}y}{d{t}^2}\propto (-y)$
Hence, it is the equation of an SHM with amplitude $A=\sqrt{a^2+b^2}$

Hence, options (a) and (d) are correct.