The differential equation of a particle performing S.H.M. is d t 2 d 2 x + 64 x = 0 , where x is in meter. The period of oscillation of the particle is (in second)

Question

The differential equation of a particle performing S.H.M. is d t 2 d 2 x ​ + 64 x = 0 , where x is in meter. The period of oscillation of the particle is (in second)

TestHub · Accepted Answer

The standard differential equation for a particle performing Simple Harmonic Motion is given by

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

Comparing the given equation $\frac{d ^{2} x}{d t ^{2}} + 64 x = 0$ with the standard equation, we get

$ω^{2} = 64$

$ω = 8$ rad/s

The time period of oscillation $T$ is given by

$T = \frac{2 π}{ω}$

$T = \frac{2 π}{8} = \frac{π}{4}$ s

Answer: $\frac{π}{4}$

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