Lab Report 2

Simple Harmonic Motion Alonzo Cortez PHYS 1430 Lab Section 501 TA: Soumya Sahoo 4/11/23

Alonzo Cortez Department of Physics, University of North Texas, 1155 Union Circle, #311427, Denton, Texas 76203 11th of April, 2023 Abstract For this lab, our objective is to investigate two examples of harmonic motion and be able determine which aspects of each system affects its period. Harmonic motion is the repeating periodic motion through a place. Pendulums and mass-spring systems, which are both known to have a simple harmonic motion, are used to illustrate this concept. Several sorts of balls were attached to a thread to inspect the pendulum. Then, to determine the relationship between the two, the periods were evaluated at various string lengths and amplitudes. Similar to this, a brass and steel spring rather than a string was used to test the mass-spring system. Using this, we were able to compute parameters like period length, g value, restoring force, spring constants, and frequency. Error percentages were taken, with an average 4.1 percent and a small margin of standard deviation as a result of a low range. The mass of the spring and the spring constant both had an impact on the period for the mass-spring system. When a mass is heavier, the period is longer; when a spring constant is higher, the period is slower. By understanding this lab of simple harmonic motions, we can hold a better grasp for intricate movements seen in the real world. Introduction Hooke's Law states that harmonic motion must abide by this relationship, where F is the strength of the restoring force. K is the force constant and x is the amplitude. Equation 1 1. 𝐹 = 𝑘𝑥 The expression below relates harmonic motion's period and frequency.T is the period in seconds and f is the frequency in hertz. Equation 2. 2. 𝑇 = 1/𝐹 For the pendulum, variables consist of the length of the string, the amplitude, and the mass of the pendulum bob. Variables for the mass suspended from the oscillating spring are the amplitude, the mass suspended, and the spring . You will be required to implement each of these changes and gather data throughout the procedure and data portion of this study. This data may then be evaluated to identify which of these elements has an impact on the system under consideration. You should be able to confirm the following relationships between the periods of a pendulum and a mass on

Figure 1: The above image shows Robert Hooke, who is said to have discovered the Hooke’s Law. The image above shows Rober Hooke, an English polymath said to have discovered Hooke's Law. According to the physics principle known as Hooke's law, the force needed to stretch or compress a spring is inversely correlated with the spring's displacement from its equilibrium position. The mathematical formula for Hooke's law is F = -kx, where x is the displacement from the equilibrium position, k is the spring constant, which is a measure of the stiffness of the spring, and F is the restoring force. In the study of mechanics, Hooke's law is a key idea that is applied in the construction of several mechanical components and buildings, including suspension systems, shock absorbers, and bridges. Apparatus ● Support nods and clamp ● One brass spring and one steel spring ● Balance ● Timer ● String ● Meter stick ● One wooden ball and one brass ball ● Set of masses Data, Calculations and Graphs Pendulum

Mass: Brass Ball: 66.9 g Wooden Ball: 6.8 g Pendulum Bob Length (cm) Amplitude (cm) Frequency (Hz) Period (s) % Error Brass Ball 90.2 cm 10 cm 0.55 20s 5.37 Brass Ball 90.2 cm 20 cm 0.5 20s 4.6 Brass Ball 38. cm 10 cm 0.85 20s 3.23 Wooden Ball 38 cm 10 cm 0.85 20s 3.23 Table 1: Demonstrates the table with the recorded information gathered from the pendulum. Calculation will be made in reference to the brass ball at 90.2 cm in length and 10 cm in amplitude Reference to Equation 6 Frequency = Cycles/ Time = 11/20 = 0.55 Reference to Equation 5 Percent Error = (M-A/A) x 100% = (1.05-1.81/1.81) x 100% = 5.37% Time vs Mass Time vs amp

Brass 0.32 cm 0.864 cm 0.544 cm .55 5.39 9.91 Steel 0.208 cm 0.37 cm 0.162 cm .55 5.39 32.7 All calculation will be made in reference to brass spring Reference to Equation 7 Restoring force = Mg = .55(9.8) = 5.39 Reference to Equation 1 F = kx 5.39 = k(0.54) k= 9.91 Spring Mass(g) Amplitude (cm) Frequency (hertz) Period Frequency from eq 4 % error Brass 500 g 5 cm 0.65 1.54 1.48 3.97 Brass 500 g 10 cm 0.7 1.43 1.48 3.44 Brass 1000 g 5 cm 0.5 2 2.05 2.5 Brass 500 g 5 cm 1.2 0.83 0.81 2.44 All calculations will be made in reference to brass spring at 500 g Reference to Equation 6 Frequency = Cycles/ Time = 13/20 = 0.65 Reference to Equation 3 𝑇 = 2π√𝐿⁄𝑔 = 2π 0.588/9.8 √ = 1.54 Reference to Equation 4 𝑇 = 2π√𝑚⁄𝑘 = 2 500/9011 π√ = 1.48

Time vs Mass Time vs Amplitude Time vs spring constant