A particle of mass $10 g$ is placed in potential field, given by $V = (50 x^{2} + 100) e r g / g$ . What will be frequency of oscillation of particle?

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Solution

Formula,

$\frac{d^{2} x}{d t^{2}} + w^{2} x = 0$ ........A

Given,

$V = (50 x^{2} + 100)$

$U = 10 (50 x^{2} + 100)$

$F = - \frac{d U}{d x} = - 1000 x$ .......(1)

$F = m a s s \times a c c e l e r a i o n$

$F = 10 \frac{d^{2} x}{d t^{2}}$ ..........(2)

From (1) and (2), we have,

$- 1000 x = 10 \frac{d^{2} x}{d t^{2}}$

$\frac{d^{2} x}{d t^{2}} + 100 x = 0$

From A,

$w^{2} = 100$

$n = \frac{w}{2 π} = \frac{\sqrt{100}}{2 π} = 1.59 H z$

Hence the frequency of oscillation of particle.

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