Question

Two equations of two SHM are y= asin (wt - alpha) and y = bcos (wt- alpha). Prove that phase difference beaten two the two is 90°. Pls tell the formula of phase difference and solve.

Answer 1

Explanation:

Thanks for Reading

Hope Helps You

Mark it Brainlist

Answer 2

The phase difference is 90°.

Explanation:

Given that,

Two equation of SHM is

$y_{1}=a\sin(\omega t-\alpha)$

$y_{2}=b\cos(\omega t-\alpha)$

Phase difference :

Phase difference is the difference of phase angle.

From second equation,

$y_{2}=b\cos(\omega t-\alpha)$

Phase angle $\phi =\alpha$

We know that,

$\cos\theta=\sin(\dfrac{\pi}{2}+\theta)$

Put the value of $\cos\theta$ in the equation y₂

$y_{2}=b\sin(\omega t-\alpha+\dfrac{\pi}{2})$

Here, phase angle $\phi=\dfrac{\pi}{2}+\alpha$

We need to calculate the phase difference

Using formula of phase difference

$\Delta \phi=\phi_{2}-\phi_{1}$

$\Delta \phi=\dfrac{\pi}{2}-\alpha-(-\alpha)$

$\Delta\phi=\dfrac{\phi}{2}$

$\Delta \phi=\dfrac{\pi}{2}=90^{\circ}$

Hence, The phase difference is 90°.

Learn more :

Topic : phase difference

https://brainly.in/question/10248160