Question

The equation of motion of a particle executing simple harmonic motion is a + 16 Pi Square x is equal to zero in this equation a is the linear acceleration in metre per second of the particle at a displacement x in metre find the time period.

Answer 1

we know, A particle follow simple harmonic motion when relation between linear acceleration and displacement must be $\bf{a=-\omega^2x}$....(1)

where, $\omega$ is angular frequency.

here, The equation of motion of a particle executing simple harmonic motion is a + 16π²x = 0

or, a = -16π²x ........(2)

comparing equations (1) and (2),

$\omega^2=16\pi^2$

$\implies\omega=4\pi$

but we know, $\omega=\frac{2\pi}{T}$

so, 2π/T = 4π ⇒ T = 2π/4π = 1/2 sec or 0.5 sec

hence, time period is 0.5 sec.

Answer 2

Heya mate,

Answer:

Refer to the attachment for your answer...

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