A \(200-\mathrm{g}\) mass is attached to a spring of constant \(k=5.6 \mathrm{N} / \mathrm{m}\) and set into oscillation with amplitude \(A=25 \mathrm{cm} .\) Determine (a) the frequency in hertz, (b) the period, (c) the maximum velocity, and (d) the maximum force in the spring.

Understanding the spring constant is pivotal in analyzing simple harmonic motion (SHM). The spring constant, denoted by the symbol \(k\), measures the stiffness of a spring. In more technical terms, it is the force required to extend or compress the spring by a unit length. The higher the spring constant, the stiffer the spring, meaning more force is needed for the same displacement.
In the given exercise, a spring constant of \(5.6 \, \mathrm{N/m}\) signifies that each meter of extension requires 5.6 Newtons of force. It's an intrinsic property of the material and geometry of the spring. An easy way to imagine this is by thinking of a tight guitar string compared to a loose one; the tighter string has a higher spring constant and requires more effort to displace.

The oscillation frequency of an object in SHM is the number of oscillations it makes per second. It is measured in hertz (Hz) and is symbolized by \(f\). Frequency is governed by the properties of the system, such as mass and spring constant, using the relationship \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\).

For a mass-spring system, if either the mass \(m\) increases or the spring constant \(k\) decreases, the system will oscillate less frequently, and vice versa. In the textbook exercise, after converting the mass to kilograms, the calculation of frequency involves using the spring constant and the mass to find how many times the mass will oscillate back and forth over a second.

Maximum velocity in SHM is achieved when the oscillating object passes through the equilibrium position because it is here that the object's kinetic energy is at its peak. To calculate this value, the equation \(V_{max} = A\omega\) is used, where \(A\) is the amplitude of oscillation, and \(\omega\) is the angular frequency. Angular frequency relates to linear frequency through \(\omega = 2\pi f\).

The amplitude represents the maximum displacement from the equilibrium point; thus, converting it to meters is vital before plugging it into the equation. Once you have the oscillation frequency, you multiply it by the amplitude and by \(2\pi\) to find the maximum speed the mass achieves during its motion.

The period of oscillation, symbolized as \(T\), is the time it takes for one complete cycle of oscillation. While frequency tells us how many cycles occur in a second, the period tells us the duration of one such cycle in seconds. The period and frequency are inversely related, which can be expressed as \(T = \frac{1}{f}\).

After determining the frequency, its reciprocal gives the period in seconds. It's essential for predicting when the object will return to a particular point in its oscillatory path. For instance, if a clock's pendulum has a period of 2 seconds, we can anticipate it will take 2 seconds to swing back to the starting point after each swing.