A particle executes linear simple harmonic motion with amplitude of 2

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:

2π

2π/√3

√3/2π

2π√3

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-2 – 2016

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For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $\omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = \pm \omega \sqrt{A^2 – x^2}$, and the acceleration $a$ is given by $a = -\omega^2 x$. The magnitudes are $|v| = \omega \sqrt{A^2 – x^2}$ and $|a| = \omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$.
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 $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{\sqrt{3}}$.

Formulas for velocity and acceleration in SHM are $|v| = \omega \sqrt{A^2 – x^2}$ and $|a| = \omega^2 |x|$. The time period is $T = 2\pi/\omega$.

The velocity is maximum at the mean position ($x=0$) and zero at the extreme positions ($x=\pm A$). The acceleration is maximum at the extreme positions and zero at the mean position. The relationship between velocity and displacement is elliptical in phase space, while the relationship between acceleration and displacement is linear.

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