Question

Two $SHM$ are represented by equation $x_1=4\sin (\omega t+{37}^o)$ and $x_2=5\cos \left(\omega t\right)$. The phase difference between them is

• A. ${37}^o$
• B. ${127}^o$
• C. ${53}^o$
• D. ${143}^o$

Answer 1

The correct option is C ${53}^o$

Given,

$x=A\sin (\omega t+\varphi ),$

$where,$

$A=$ Amplitude

$\varphi =$ Phase Angle

$x_1=4\sin (\omega t+{37}^o)\dots \dots \left(1\right)$

$x_2=5\cos \left(\omega t\right)=5\sin (\omega t+{90}^o)\dots \dots \left(2\right)$

Phase angle difference is ${90}^o-{37}^o={53}^o$