A particle executes simple harmonic motion according to equation $4 \frac{d^2 x}{d t^2}+320 x=0$. Its time period of oscillation is .........
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(c)
$4 \frac{d^2 x}{d t^2}+320 x=0$
$4 a=-320 x$
$a=-80 x$
Since $a=-\omega^2 x$ in $S.H.M.$
$80=\omega^2$
$\sqrt{16 \times 5}=\omega$
or $\omega=4 \sqrt{5}$
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{4 \sqrt{3}}=\frac{\pi}{2 \sqrt{5}} s$
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