A particle executes linear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is:
[amp_mcq option1=”2π” option2=”2π/√3″ option3=”√3/2π” option4=”2π√3″ correct=”option2″]
This question was previously asked in
UPSC NDA-2 – 2016
Substituting the values: $\omega \sqrt{2^2 – 1^2} = \omega^2 |1|$
$\omega \sqrt{4 – 1} = \omega^2$
$\omega \sqrt{3} = \omega^2$
Since $\omega$ for SHM is non-zero, we can divide by $\omega$:
$\sqrt{3} = \omega$.
The time period $T$ is related to angular frequency $\omega$ by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{\sqrt{3}}$.