For a periodic motion represented by the equation y = sin ωt + cos ωt, the amplitude of the motion is

Question

Two vectors are given by $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{B} = 2\hat{i} - 4\hat{j} + 3\hat{k}$. The unit vector along $\vec{A} + \vec{B}$ is

Answer 1

To find the amplitude of the given periodic motion represented by the equation $ y = \sin \omega t + \cos \omega t $, we can utilize a trigonometric identity to simplify the expression.

The expression $ y = \sin \omega t + \cos \omega t $ can be treated as a resultant of two perpendicular harmonic motions. We need to express this sum in the form of a single sinusoidal function $ y = A \sin(\omega t + \phi) $, where $ A $ is the amplitude.

The formula for the amplitude $ A $ of the function of the form $ y = a \sin \omega t + b \cos \omega t $ is given by:

A = \sqrt{a^2 + b^2}

In our equation:

$ a = 1 $ (coefficient of $\sin \omega t$)
$ b = 1 $ (coefficient of $\cos \omega t$)

Substitute these values into the formula:

A = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}

Thus, the amplitude of the motion represented by the equation $ y = \sin \omega t + \cos \omega t $ is $ \sqrt{2} $.

Therefore, the correct answer is:

$\sqrt{2}$

This result is consistent with the given correct answer.

Answer 2