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

\frac{1}{2 π} \sqrt{\frac{b}{a}}

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\frac{\sqrt{a}}{2 π}

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\frac{\sqrt{b}}{2 π}

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\frac{1}{2 π} \sqrt{\frac{a}{b}}

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Solution

The correct option is B $\frac{\sqrt{a}}{2 π}$
$v = \sqrt{b - a x^{2}} = \sqrt{a (\frac{b}{a} - x^{2})} = \sqrt{a} \times \sqrt{⎛ ⎝ {(\sqrt{\frac{b}{a}})}^{2} - x^{2} ⎞ ⎠} - - - - - (1)$
Formula for velocity is given by,
$v = ω \times \sqrt{A^{2} - x^{2}} - - - - - (2) ω = \sqrt{a} 2 π f = \sqrt{a} ∴ f = \frac{\sqrt{a}}{2 π}$

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