A particle executes linear simple harmonic motion with an

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Given, Amplitude A = 3 cm
When particle is at x = 2 cm
According to question, magnitude of velocity = acceleration
ω √A² - x² = x ω²

√3² - 2² = 2		2π
		T

√5 =	4π	⇒ T =	4π
	T		√5

	2π
	√3