Given:

Time for one complete swing (period) $T = 0.1$ seconds
The spring is not stretched when the mass is at its highest point (this is the farthest point from the middle position).
Gravitational acceleration, $g = 10\, \mathrm{m/s}^2$

Let's find the answer step by step.

Step 1: Formula for Maximum Speed in S.H.M.

In simple harmonic motion, the highest speed is given by:

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

Where:

$A$ is the amplitude (the maximum distance from the center position)
$\omega$ is the angular frequency (tells us how fast the object goes in a circle or swings back and forth)

Step 2: Find Angular Frequency $\omega$

The formula connecting period and angular frequency is:

$$ \omega = \frac{2\pi}{T} $$

Plug in $T = 0.1$:

$$ \omega = \frac{2\pi}{0.1} = 20\pi $$

Step 3: Find Amplitude $A$

At the highest position, the spring is just relaxed (not stretched). This point is $A$ distance from the mean (middle) position.

At the middle (mean position), gravity balances the pull of the spring: $mg = kx_0$, where $x_0$ is the stretch from natural length.
At highest point: extension is $0$ (spring just relaxed). This distance from the middle to the top is our amplitude $A$.
So, $A = x_0$.

Thus, to find $A$ we find $x_0$.

From $mg = kx_0$:

$$ x_0 = \frac{mg}{k} $$

Step 4: Connect $k$ and $m$ to $\omega$

For a spring and mass, angular frequency is:

$$ \omega = \sqrt{\frac{k}{m}} $$

This means $ \frac{k}{m} = \omega^2 $.

So, substitute into earlier expression for $x_0$:

$$ x_0 = \frac{mg}{k} = \frac{g}{k/m} = \frac{g}{\omega^2} $$

Step 5: Plug Amplitude into the Speed Formula

Recall:

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

Since $A = x_0 = \frac{g}{\omega^2}$, then:

$$ v_{max} = \frac{g}{\omega^2} \cdot \omega = \frac{g}{\omega} $$

Step 6: Substitute the Values

Now calculate maximum speed:

$$ v_{max} = \frac{10}{20\pi} = \frac{1}{2\pi} \ \mathrm{m/s} $$

Step 7: Final Answer

The correct answer is:

$$ \boxed{\frac{1}{2\pi} \,\mathrm{m/s}} $$

Simple Harmonic Motion

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