The velocity vector v and displacement vector x of a particle executing SHM are related as Vdvdx-w2x with the initial condition v-v0 at x-0- The velocity vs when displacement is x- is

The correct option is B v=√( v _ 0 ^ 2 w ^ 2 x ^ 2 ) Hence the correct option is B. #160; ...

You visited us 7 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Fundamental and Derived Units

The velocity ...

Question

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $V \frac{d v}{d x} = - w^{2} x$ with the initial condition $v = v_{0}$ at $x = 0$ . The velocity $v_{s}$ when displacement is $x$ , is

v = \sqrt{v_{0}^{2} + w^{2} x^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = \sqrt{v_{0}^{2} - w^{2} x^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

v = \sqrt{v_{0}^{3} + w^{3} + x^{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = v_{0} - {(w^{3} x^{3} e^{x^{3}})}^{1 / 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $v = \sqrt{v_{0}^{2} - w^{2} x^{2}}$

Suggest Corrections

Similar questions

The velocity v and displacement x of a particle executing simple harmonic motion are related as

$v \frac{d v}{d x} = - ω^{2} x$

At $x = 0, v = v_{0}$ . Find the velocity v when the displacement becomes x.

The velocity v and displacement x of a particle executing simple harmonic motion are related as

$v \frac{d v}{d x} = - ω^{2} x$

At $x = 0, v = v_{0}$ . Find the velocity v when the displacement becomes x.

Q. A particle is projected under gravity with a speed

v_{0}

. Using

W-E

theorem, find the speed of the particle as the function of vertical distance

y

Q. If the displacement

x

and velocity

v

of a particle executing S.H.M are related through the expression

4 v^{2} = 25 - x^{2}

, then its maximum displacement in meters is

If the displacement $x$ and velocity $v$ of a particle executing simple harmonic motion are related through the expression $4 v^{2} = 25 - x^{2}$ , then its time period is

Join BYJU'S Learning Program

The velocity vector v and displacement vector x of a particle executing SHM are related as Vdvdx=−w2x with the initial condition v=v0 at x=0. The velocity vs when displacement is x, is

The correct option is B v=√v20−w2x2Hence the correct option is B.

Hence the correct option is B.

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $V \frac{d v}{d x} = - w^{2} x$ with the initial condition $v = v_{0}$ at $x = 0$ . The velocity $v_{s}$ when displacement is $x$ , is

The correct option is B $v = \sqrt{v_{0}^{2} - w^{2} x^{2}}$