A particle is executing S.H.M. and its velocity v is related to its position (x) as v2+a x2=b, where a and b are positive constants. The frequency of oscillation of particle is

Question

A particle is executing S.H.M. and its velocity v is related to its position (x) as v2+a x2=b, where a and b are positive constants. The frequency of oscillation of particle is (A) (1/2 π) √(b/a) (B) (√a/2 π) (C) (√b/2 π) (D) (1/2 π) √(a/b)

Tardigrade Admin · Accepted Answer

v2+a x2=b v2=b-a x2 v2=a((b/a)-x2) Comparing it to equation v2=ω2(A2-x2) ω=√a f=(ω/2 π)=(√a/2 π)

Q. A particle is executing S.H.M. and its velocity $v$ is related to its position $(x)$ as $v^{2} + a x^{2} = b$ , where $a$ and $b$ are positive constants. The frequency of oscillation of particle is

$\frac{1}{2 π} \frac{b}{a}$

$\frac{a}{2 π}$

$\frac{b}{2 π}$

$\frac{1}{2 π} \frac{a}{b}$

$v^{2} + a x^{2} = b$
$v^{2} = b - a x^{2}$
$v^{2} = a (\frac{b}{a} - x^{2})$
Comparing it to equation
$v^{2} = ω^{2} (A^{2} - x^{2})$
$ω = a$
$f = \frac{ω}{2 π} = \frac{a}{2 π}$

Q. A particle is executing S.H.M. and its velocity v is related to its position (x) as v2+ax2=b, where a and b are positive constants. The frequency of oscillation of particle is

2π1​ab​​

2πa​​

2πb​​

2π1​ba​​

v2+ax2=b v2=b−ax2 v2=a(ab​−x2) Comparing it to equation v2=ω2(A2−x2) ω=a​ f=2πω​=2πa​​