mum speed.

Determine Angular Frequency ($\omega$):

The angular frequency $\omega$ is related to the frequency $f$ by the formula:

$$ \omega = 2\pi f $$

Given $ f = 10 \, \text{Hz} $, we have:

$$ \omega = 2\pi \times 10 = 20\pi \, \text{rad/s} $$

Maximum Speed Formula in SHM:

The maximum speed $v_{\text{max}}$ of a particle in SHM is given by:

$$ v_{\text{max}} = A \omega $$

where $A$ is the amplitude of the oscillation.

Determine Amplitude ($A$):

At the highest point, the spring is unstretched, meaning the elongation due to the particle's weight is equal to the amplitude of oscillation.

Use the equilibrium condition $mg = kA$, and since $ \omega = \sqrt{\frac{k}{m}} $, $k = m\omega^2$.

Therefore, at maximum elongation, the net force provides the centripetal force for the circular motion:

$$ mg = m\omega^2 A $$

Consequently, $ A = \frac{g}{\omega^2} $.

Substitute Values to Find $v_{\text{max}}$:

Substituting for amplitude:

$$ A = \frac{10}{(20\pi)^2} = \frac{10}{400\pi^2} = \frac{1}{40\pi^2} $$

Now, substitute $A$ and $\omega$ into $v_{\text{max}} = A \omega$:

$$ v_{\text{max}} = \left(\frac{1}{40\pi^2}\right) \times 20\pi = \frac{20\pi}{40\pi^2} $$

$$ v_{\text{max}} = \frac{1}{2\pi} \, \text{m/s} $$

Thus, the maximum speed of the particle is $\frac{1}{2\pi} \, \text{m/s}$.

Answer:

Option B: $\frac{1}{2 \pi} \, \text{m/s}$

Simple Harmonic Motion

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