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Kinetic energy of a particle of mass 40 gm executing simple harmonic motion along a straight line is given by $K = 16 (1 - 0 .25 x^{2})$ joule where x is in cm. Then frequency of oscillation of the particle is

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$\frac{500 \sqrt{2}}{π} Hz$

$\frac{500}{π} Hz$

$\frac{250}{π} Hz$

$\frac{1000}{π} Hz$

answer is A.

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Detailed Solution

Kinetic energy $K = \frac{1}{2} {mω}^{2} A^{2} (1 - \frac{x^{2}}{A^{2}})$

Comparing this equation with the given equation we get, $A^{2} = \frac{1}{0 .25} \Rightarrow A = 2 cm$

and $\frac{1}{2} {mω}^{2} A^{2} = 16 \Rightarrow \frac{1}{2} \times \frac{40}{1000} . ω^{2} . {(\frac{2}{100})}^{2} = 16$

$\Rightarrow ω^{2} = 16 \times 50 \times 2500 ω = 1000 \sqrt{2} rad / s$

$∴ f = \frac{ω}{2 π} = \frac{500 \sqrt{2}}{π} Hz$

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