Two vectors vec{A} and vec{B} are defined as | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\vec{A} = a \hat{i}$ and $\vec{B} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.$\vec{A} + \vec{B} = a(1 + \cos \omega t) \hat{i} +

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $\vec{A} = a \hat{i}$ and $\vec{B} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.$\vec{A} + \vec{B} = a(1 + \cos \omega t) \hat{i} + a \sin \omega t \hat{j}$.$|\vec{A} + \vec{B}|^2 = a^2(1 + \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 + 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 + 2 \cos \omega t) = 2a^2(1 + \cos \omega t) = 4a^2 \cos^2(\omega t / 2)$.So,$|\vec{A} + \vec{B}| = 2a \cos(\omega t / 2)$.Similarly,$\vec{A} - \vec{B} = a(1 - \cos \omega t) \hat{i} - a \sin \omega t \hat{j}$.$|\vec{A} - \vec{B}|^2 = a^2(1 - \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 - 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 - 2 \cos \omega t) = 4a^2 \sin^2(\omega t / 2)$.So,$|\vec{A} - \vec{B}| = 2a \sin(\omega t / 2)$.Given $|\vec{A} + \vec{B}| = \sqrt{3}|\vec{A} - \vec{B}|$,we have $2a \cos(\omega t / 2) = \sqrt{3} \cdot 2a \sin(\omega t / 2)$.$	an(\omega t / 2) = 1 / \sqrt{3}$.$\omega t / 2 = \pi / 6 \implies \omega t = \pi / 3$.Since $\omega = \pi / 6 	ext{ rad s}^{-1}$,we get $(\pi / 6) \cdot t = \pi / 3$.Therefore,$t = 2 	ext{ s}$.

Similar Questions

The vector sum of two forces is perpendicular to their vector difference. In that case,the forces

Explain the subtraction of vectors.

Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^{\circ}$,then which of the following vectors may be equal to $\frac{\vec A}{2} - \vec B$?

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

Two equal forces ($P$ each) act at a point inclined to each other at an angle of $120^\circ$. The magnitude of their resultant is

Vedclass Test Series

Exam Paper Generator

Online Exam Module