Frequency of a System Performing Simple Harmonic Motion (SHM)

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. The frequency of a system undergoing SHM is a fundamental characteristic that describes how often the motion repeats itself in a given time period.

Understanding Frequency in SHM

Frequency, denoted by the symbol $f$, is defined as the number of oscillations or cycles per unit time. The unit of frequency is Hertz (Hz), where 1 Hz equals one cycle per second.

The formula for frequency in terms of the period $T$ (the time taken to complete one cycle) is:

$$ f = \frac{1}{T} $$

Determinants of Frequency in SHM

The frequency of a system performing SHM depends on the properties of the system, such as mass and stiffness in the case of a mass-spring system, or length and gravitational acceleration for a simple pendulum.

For a Mass-Spring System:

The frequency of a mass-spring system is given by:

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

where:

• $k$ is the spring constant (measure of stiffness),
• $m$ is the mass of the object attached to the spring.

For a Simple Pendulum:

The frequency of a simple pendulum is given by:

$$ f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} $$

where:

• $g$ is the acceleration due to gravity,
• $L$ is the length of the pendulum.

Table: Differences in Determinants of Frequency

System Type	Determinants of Frequency	Formula for Frequency
Mass-Spring System	Mass ($m$), Spring constant ($k$)	$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
Simple Pendulum	Length ($L$), Gravitational acceleration ($g$)	$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$

Examples

Example 1: Mass-Spring System

Consider a mass-spring system with a mass of 0.5 kg and a spring constant of 200 N/m. The frequency of the system is calculated as follows:

$$ f = \frac{1}{2\pi} \sqrt{\frac{200 \text{ N/m}}{0.5 \text{ kg}}} \approx \frac{1}{2\pi} \sqrt{400 \text{ s}^{-2}} \approx \frac{20}{2\pi} \text{ Hz} \approx 3.18 \text{ Hz} $$

Example 2: Simple Pendulum

For a simple pendulum with a length of 1 meter and assuming $g = 9.81 \text{ m/s}^2$, the frequency is:

$$ f = \frac{1}{2\pi} \sqrt{\frac{9.81 \text{ m/s}^2}{1 \text{ m}}} \approx \frac{1}{2\pi} \sqrt{9.81 \text{ s}^{-2}} \approx \frac{3.13}{2\pi} \text{ Hz} \approx 0.498 \text{ Hz} $$

Conclusion

The frequency of a system performing SHM is a crucial parameter that depends on the physical properties of the system. For a mass-spring system, it depends on the mass and the spring constant, while for a pendulum, it depends on the length and the gravitational acceleration. Understanding these relationships allows us to predict the behavior of oscillatory systems and design them for specific applications.