Question

A particle is executing SHM $x=3\cos \omega t+4\sin \omega t$. Find the phase shift and amplitude.

• A. ${50}^{\circ },5$ units
• B. ${37}^{\circ },3.5$ units
• C. ${53}^{\circ },3.5$ units
• D. ${37}^{\circ },5$ units

Answer 1

The correct option is D ${37}^{\circ },5$ units

$x=3\cos \omega t+4\sin \omega t$
lets take
$x_0\cos \varphi =4$ ............$\left(I\right)$
and $x_0\sin \varphi =3$ ............$\left(II\right)$
Now the equation becomes
$x=x_0\sin \varphi \cos \omega t+x_0\cos \varphi \sin \omega t\Rightarrow x=x_0(\sin \varphi \cos \omega t+\cos \varphi \sin \omega t)\Rightarrow x=x_0\sin (\omega t+\varphi )$
Squaring and adding $\left(I\right)$ and $\left(II\right)$ we get
$\Rightarrow {x_0}^2({\cos }^2\varphi +{\sin }^2\varphi )=4^2+3^2=25\Rightarrow x_0=5$
dividing $\left(II\right)$ by $\left(I\right)$, we get
$\tan \varphi =\frac{3}{4}\Rightarrow \varphi ={37}^{\circ }$
$\Rightarrow x=5\sin (\omega t+{37}^{\circ })$ which represents a SHM of amplitude of 5 units and a phase of ${37}^{\circ }$