Diver Mass and Springboard Physics

This problem involves understanding the principles of oscillations in a mechanical system. We are asked to determine the static deflection of a spring board when a diver of a specific mass is standing on it. We are provided with the natural frequency of the system, which includes the diver and the spring board.

Identified Diver and Springboard Values

Let's identify the given parameters from the question:

• Mass of the diver, denoted as $m = 100$ kg.
• Natural frequency of the diver-springboard system, denoted as $f = 1.6$ Hz.
• The mass of the spring board itself is negligible.
• We will use the standard acceleration due to gravity, $g \approx 9.81$ m/s$^2$.

Natural Frequency, Mass, and Spring Constant Relationship

For a mass-spring system undergoing simple harmonic motion, the natural frequency is directly related to the mass and the stiffness of the spring (spring constant). The key relationships are:

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

$$ \omega = 2 \pi f $$

The angular frequency is also defined in terms of mass, $m$, and spring constant, $k$, as:

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

By equating these two expressions, we can derive the relationship between frequency, mass, and spring constant:

$$ 2 \pi f = \sqrt{\frac{k}{m}} $$

Squaring both sides to solve for $k$:

$$ (2 \pi f)^2 = \frac{k}{m} $$

$$ 4 \pi^2 f^2 = \frac{k}{m} $$

Thus, the spring constant $k$ can be calculated using:

$$ k = 4 \pi^2 m f^2 $$

Spring Constant Calculation Step

Using the identified values, we can calculate the spring constant $k$ of the spring board system:

$$ k = 4 \pi^2 (100 \text{ kg}) (1.6 \text{ Hz})^2 $$

$$ k = 4 \pi^2 (100) (2.56) \text{ N/m} $$

$$ k = 1024 \pi^2 \text{ N/m} $$

Static Deflection Calculation Step

Static deflection ($\delta$) is the displacement of the spring from its equilibrium position when a constant force is applied. This force is the weight of the diver, $F = mg$. According to Hooke's Law, $F = k \delta$. Therefore, the static deflection is:

$$ \delta = \frac{F}{k} $$

Substituting $F = mg$ and the expression for $k$ ($k = 4 \pi^2 m f^2$):

$$ \delta = \frac{mg}{4 \pi^2 m f^2} $$

The mass $m$ cancels out in this equation:

$$ \delta = \frac{g}{4 \pi^2 f^2} $$

Deflection Calculation and Unit Conversion

Now, we substitute the known values ($g = 9.81$ m/s$^2$ and $f = 1.6$ Hz) into the formula for static deflection:

$$ \delta = \frac{9.81 \text{ m/s}^2}{4 \pi^2 (1.6 \text{ Hz})^2} $$

$$ \delta = \frac{9.81}{4 \pi^2 (2.56)} \text{ m} $$

$$ \delta = \frac{9.81}{10.24 \pi^2} \text{ m} $$

Using $\pi \approx 3.14159$ (so $\pi^2 \approx 9.8696$):

$$ \delta \approx \frac{9.81}{10.24 \times 9.8696} \text{ m} $$

$$ \delta \approx \frac{9.81}{101.068} \text{ m} $$

$$ \delta \approx 0.09706 \text{ m} $$

To express the deflection in millimeters, we multiply the result by 1000:

$$ \delta \approx 0.09706 \text{ m} \times 1000 \text{ mm/m} $$

$$ \delta \approx 97.06 \text{ mm} $$

Conclusion on Static Deflection

Our calculation yields a static deflection of approximately 97.06 mm. Comparing this value with the provided options, 98.1 mm is the closest option. Thus, the static deflection at the tip of the spring board when the diver is standing is 98.1 mm.