91Ó°ÊÓ

In simple harmonic motion (SHM), a mass-spring system is a common example. SHM describes the periodic back-and-forth motion of an object. The object will experience a force that tries to bring it back to the equilibrium position.

In our case, we have a small body of mass, which is connected to a spring. This setup helps us visualize and understand SHM easily:

• Mass (m): 0.12 kg
• Amplitude (A): 8.5 cm or 0.085 m
• Period (T): 0.20 s

The amplitude is the maximum displacement from the equilibrium position. The period is the time it takes for one complete cycle of motion. When the mass is displaced, the spring exerts a force proportional to the displacement, always trying to restore the mass back to equilibrium.

As the mass moves, it oscillates between the maximum displacement in either direction. The whole motion is very predictable and can be described using mathematical formulas.

The maximum force in a mass-spring system can be found by first calculating the maximum acceleration of the system. This acceleration happens when the mass is at maximum displacement.

The formula for maximum force is:
\( F_{max} = m \, a_{max} \)
Here, we need to find \( a_{max} \), which can be obtained using the angular frequency \( \omega \).

Angular frequency is given by:
\( \omega = \frac{2\pi}{T} \)
For our system:
\( \omega = \frac{2 \, 3.1416}{0.20} = 31.42 \, \text{rad/s} \)

Then, the maximum acceleration is:
\( a_{max} = \omega^2 \, A \)
Substitute the values:
\( a_{max} = (31.42)^2 \, 0.085 = 84.36 \, \text{m/s}^2 \)
Finally, find the maximum force:
\( F_{max} = 0.12 \, 84.36 = 10.12 \, \text{N} \)

This force represents the largest force exerted on the mass by the spring during its oscillation.

To determine the spring constant (k) of the spring in our mass-spring system, we use the relationship between the period of the oscillation and the characteristics of the mass and spring.

The period (T) of oscillation is related to the mass (m) and the spring constant (k) by the following formula:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Rearrange the equation to solve for \( k \):
\( k = \frac{4\pi^2 m}{T^2} \)
Plug in the given values:
\( k = \frac{4 \ (3.1416)^2 \ 0.12}{(0.20)^2} = 118.4 \ \text{N/m} \)

The spring constant tells us how stiff the spring is. A higher value signifies a stiffer spring. In this system, the spring constant of 118.4 N/m means that the spring exerts a force of 118.4 N for each meter of stretch or compression.