A particle executes simple harmonic motion according to equation … - Vidyadip

MCQ

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 .........

A
$\frac{2 \pi}{5 \sqrt{3}} s$
B
$\frac{\pi}{3 \sqrt{2}} s$
✓
$\frac{\pi}{2 \sqrt{5}} s$
D
$\frac{2 \pi}{\sqrt{3}} s$

✓

Answer

Correct option: C.

$\frac{\pi}{2 \sqrt{5}} s$

c
(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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