Physics 1: Detailed Analysis of Mass Spring Systems Natural Frequency

This physics assignment focuses on the analysis of mass-spring systems and the calculation of their natural frequency. The solution begins by explaining the basic concepts of mass-spring systems, including the role of springs, mass, and dampers. The assignment highlights that the natural frequency is the frequency at which the system oscillates without any external forces. The solution then presents the formula for calculating the natural frequency and applies it to a specific example, providing a step-by-step calculation. The assignment also provides the calculated natural frequency (0.6495 Hz) and references relevant sources such as Jarrousse (2014) and Lumen Learning (2020). The document aims to help students understand the relationship between mass, spring constant, and natural frequency in the context of a physics problem.

Running Head: PHYSICS
0
Mass Spring Systems
Natural Frequency
(Student Details: )
3/27/2020

PHYSICS 1
Mass Spring Systems
Part 1:
(a) In the mass-spring system, the mass always acts on the spring which compresses as
well as stores the energy. In this way, the spring releases above energy back towards
the mass. When the process occurs, the mass will be jumped and bounced back into
the air, over and over, unless energy is dissipated by any friction. Hence, the damper
is used as a device which resists the springs’ motion in the opposite direction
(Jarrousse, 2014).

PHYSICS 2
Source: (Lumen Learning, 2020)
On the other hand, the natural frequency of vibration of the mass-spring system in the
absence of the opposing force of the damper (Jarrousse, 2014). The natural frequency
is the frequency that gets to a point in which the system resonates. Therefore, in the
absence of a damper, the mass will continue oscillating at a specific frequency termed
as natural frequency (Lumen Learning, 2020).
In order to determine the natural frequency of the given mass-spring system:
Formula (Jarrousse, 2014):
F= ω /2π
Here, ω = √(k / m)
Thus, natural frequency: f = √(k / m) ÷ 2π
f = √(12500 N/m / 750 kg) ÷ 2π
f = √(16.66 s−2) ÷ 2π
f =4.081 ÷ 2π
f = 0.6495 Hz
Thus, for the given case, the natural frequency is 0.6495 Hz, which means the system
would oscillate just over one and a half times per second.

PHYSICS 3
References
Jarrousse, O., 2014. Modified mass-spring system for physically based deformation modeling.
14th ed. UK: KIT Scientific Publishing.
Lumen Learning, 2020. Energy in Simple Harmonic Motion. [Online] Available at:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/15-2-energy-in-simple-
harmonic-motion/ [Accessed 2020].