The velocity vector v and displacement vector x of a particle executing SHM are related as (v d v/d x)=-ω2 X with the initial condition V=v0 at x=0. The velocity v, when displacement is X, is

Question

The velocity vector v and displacement vector x of a particle executing SHM are related as (v d v/d x)=-ω2 X with the initial condition V=v0 at x=0. The velocity v, when displacement is X, is (A) v=√v02+ω2 x2 (B) v=√v02-ω2 x2 (C) v=√[3]v03+(1)3 x3 (D) v=v0- (ω3 x3ex3)1/3

Tardigrade Admin · Accepted Answer

Given, v (d v/d x)=-ω2 x On integrating within the limit ∫ limitsV0V v d v =∫ limits0x-ω2 x d x ⇒ [(v2/2)]V0v=-ω2[(x2/2)]0x ⇒ v2-v02 =-ω2 x2 ⇒ v=√v02-ω2 x2

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

$v = v_{0}^{2} + ω^{2} x^{2}$

$v = v_{0}^{2} - ω^{2} x^{2}$

$v = 3 v_{0}^{3} + (1)^{3} x^{3}$

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

Given,
$v \frac{d v}{d x} = - ω^{2} x$
On integrating within the limit
$V_{0} \int V v d v = 0 \int x - ω^{2} x d x$
$\Rightarrow [\frac{v ^{2}}{2}]_{V_{0}}^{v} = - ω^{2} [\frac{x ^{2}}{2}]_{0}^{x}$
$\Rightarrow v^{2} - v_{0}^{2} = - ω^{2} x^{2}$
$\Rightarrow v = v_{0}^{2} - ω^{2} x^{2}$

Q. The velocity vector v and displacement vector x of a particle executing SHM are related as dxvdv​=−ω2X with the initial condition V=v0​ at x=0. The velocity v, when displacement is X, is

v=v02​+ω2x2​

v=v02​−ω2x2​

v=3v03​+(1)3x3​

v=v0​−(ω3x3ex3)1/3

Given, vdxdv​=−ω2x On integrating within the limit V0​∫V​vdv=0∫x​−ω2xdx ⇒[2v2​]V0​v​=−ω2[2x2​]0x​ ⇒v2−v02​=−ω2x2 ⇒v=v02​−ω2x2​