Two vectors A → a n d B → are defined as A → = a i ^ a n d B → = a cos ⁡ ω t i ^ + sin ⁡ ω t j ^ , where a is a constant and ω = π 6 r a d s - 1 . If A → + B → = 3 A → - B → at time t = τ for the first time, the value of τ in seconds, is_______

A → + B → = 2 a cos ⁡ ω t 2 A → - B → = 2 a sin ⁡ ω t 2 So 2 a cos ⁡ ω t 2 = 3 2 a sin ⁡ ω t 2 tan ⁡ ω t 2 = 1 3 ω t 2 = π 6 ⇒ ω t = π 3 π 6 t = π 3 t = 2.00 s e c

Start Free Mock LoginAlready registered? Login here

Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousEasy2 minPYQ_2018

PhysicsEasynumerical

Two vectors $\vec{A} a n d \vec{B}$ are defined as $\vec{A} = a \hat{i} a n d \vec{B} = a (\cos ω t \hat{i} + \sin ω t \hat{j})$ , where a is a constant and $ω = \frac{π}{6} r a d s^{- 1} .$ If $|\vec{A} + \vec{B}| = \sqrt{3} |\vec{A} - \vec{B}|$ at time $t = τ$ for the first time, the value of $τ$ in seconds, is_______

Answer:

2.00

Solution:

$|\vec{A} + \vec{B}| = 2 a \cos \frac{ω t}{2}$
$|\vec{A} - \vec{B}| = 2 a \sin \frac{ω t}{2}$
So $2 a \cos \frac{ω t}{2} = \sqrt{3} (2 a \sin \frac{ω t}{2})$
$\tan \frac{ω t}{2} = \frac{1}{\sqrt{3}}$
$\frac{ω t}{2} = \frac{π}{6} \Rightarrow ω t = \frac{π}{3}$
$\frac{π}{6} t = \frac{π}{3}$
$t = 2.00 s e c$

Stream:JEE_ADVSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2018

Doubts & Discussion

Loading discussions...