The equation of SHM of a particle is given as $2 \dfrac{d^2 x}{dt^2} + 32x = 0$ where x is the displacement from the mean position. The period of oscillation (in seconds) is -

Hint: Reduce the given expression in the form of standard differential equation of SHM. Compare the two to and find the frequency of oscillation then use the relation between frequency and time period to find the time period.

Formula used:
The standard SHM differential equation is of the form:
$\dfrac{d^2 x}{dt^2} + \omega^2 x = 0$
where $\omega$ is the frequency of oscillation.

Complete step by step answer:
First we write the given expression:
$2 \dfrac{d^2 x}{dt^2} + 32x = 0$ .
Dividing both sides by 2, we get:
$\dfrac{d^2 x}{dt^2} + 16x = 0$ .
Comparison of this expression with standard differential equation for simple harmonic motion gives us:
$\omega^2 = 16$
$\omega = 4$ rad/s.
But we require to find the time period for the motion, we know that the oscillation frequency and time period are related as:
$\omega = \dfrac{2 \pi}{T}$ ,
keeping here the value of $\omega$ obtained previously we get:
$4 = \dfrac{2 \pi}{T}$
$T = \dfrac{\pi}{2}$ s.
Therefore, the time period of oscillation for the given particle is 1.57 seconds approximately.

Additional information:
Simple harmonic motion is also performed by spring undergoing small oscillations horizontally.
Derivation for standard SHM differential equation:
We know, for a spring,
F = ma = -kx
Also,
$a = \dfrac{d^2 x}{dt^2}$
We may write:
$m \dfrac{d^2 x}{dt^2} + kx = 0$ .
If we divide both sides by m and make a substitution of $\omega^2 = k/m$, we get the equation:
$\dfrac{d^2 x}{dt^2} + \omega^2 x = 0$ .
Therefore if one does not remember the standard differential equation one can use this trick to derive it easily.

Note:
There is a difference between angular frequency and frequency. Though in SHM we know that oscillation of the particle can be in a linear manner, the frequency in in the question related to SHM is always angular frequency which also appears in the standard differential equation.