In simple harmonic motion (SHM), the speed $v(x)$ at a displacement $x$ from the equilibrium position is related to the maximum speed $v_{\text{max}}$ and the amplitude $A$ by the following equation:

$v(x) = v_{\text{max}} \sqrt{1 - \left( \frac{x}{A} \right)^2}$

Where:

• $v(x)$ is the speed at displacement $x$,
• $v_{\text{max}}$ is the maximum speed,
• $A$ is the amplitude.

We are asked to find the displacement $x$ at which the speed is half of the maximum speed:

$v(x) = \frac{1}{2} v_{\text{max}}$

Substitute this into the equation for speed:

$\frac{1}{2} v_{\text{max}} = v_{\text{max}} \sqrt{1 - \left( \frac{x}{A} \right)^2}$

Canceling $v_{\text{max}}$ from both sides:

$\frac{1}{2} = \sqrt{1 - \left( \frac{x}{A} \right)^2}$

Now, square both sides:

$\frac{1}{4} = 1 - \left( \frac{x}{A} \right)^2$

Solving for $\left( \frac{x}{A} \right)^2$:

$\left( \frac{x}{A} \right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$

Taking the square root of both sides:

$\frac{x}{A} = \frac{\sqrt{3}}{2}$

Thus, the displacement $x$ is:

$x = A \cdot \frac{\sqrt{3}}{2}$

Given that the amplitude $A = 4.78 \, \text{cm}$:

$x = 4.78 \, \text{cm} \times \frac{\sqrt{3}}{2} \approx 4.78 \, \text{cm} \times 0.866$

$x \approx 4.14 \, \text{cm}$

Therefore, the displacement at which the speed is half of the maximum speed is approximately 4.14 cm.

Would you like further explanation on this or have any questions about the process?

Related Questions:

1. How does the displacement in SHM affect the speed of the particle?
2. What is the relationship between kinetic energy and displacement in SHM?
3. How would the result change if the amplitude were doubled?
4. What is the equation for the maximum speed in SHM?
5. How does the phase of SHM influence the velocity at a given point?

Tip:

In SHM, remember that the speed is maximum at the equilibrium position (where $x = 0$) and decreases as the displacement increases toward the amplitude.