Question

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

• A. \(\frac {\sqrt 5}{\pi}\)
• B. \(\frac {\sqrt 5}{2\pi}\)
• C. \(\frac {4\pi}{\sqrt 5}\)
• D. \(\frac {2\pi}{\sqrt 3}\)

Answer 1

The Correct Option is C

To solve this problem, we will use the equations of motion for a particle undergoing simple harmonic motion (SHM). The amplitude of the motion is given as 3 cm, and we need to determine the time period of the motion.

Let's start by recalling the basic equations for a particle in SHM:

• The displacement from the mean position at any time \( t \) is given by: x = A \cos(\omega t + \phi).
• The velocity in SHM is given by: v = \omega \sqrt{A^2 - x^2}.
• The acceleration in SHM is given by: a = -\omega^2 x.

According to the problem, when the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Let's derive this condition mathematically:

1. Given \( x = 2 \) cm, the velocity is: v = \omega \sqrt{3^2 - 2^2} = \omega \sqrt{5}.
2. The acceleration is: a = -\omega^2 \times 2 = -2\omega^2.
3. We know \( |v| = |a| \Rightarrow \omega \sqrt{5} = 2\omega^2 \).

Solving the equation \( \omega \sqrt{5} = 2\omega^2 \):

• \( \omega \sqrt{5} = 2\omega^2 \)
• Dividing through by \( \omega \) (assuming \( \omega \neq 0 \)): \(\sqrt{5} = 2\omega\)
• Thus, \(\omega = \frac{\sqrt{5}}{2}\).

Now, the time period \( T \) is related to \( \omega \) by the formula \( T = \frac{2\pi}{\omega} \).

• Substituting the value of \(\omega\): \( T = \frac{2\pi}{\frac{\sqrt{5}}{2}} = \frac{4\pi}{\sqrt{5}}\).

Thus, the time period of the particle is \(\frac{4\pi}{\sqrt{5}}\) seconds.