Description
experiment 3: Simple Harmonic Motion-Hooke’s Law
I need answers for the attached lab questions/experiment questions, calculation, and graphs.
Explanation & Answer
Hello buddy! I have attached the answer in a word document for you. Also, there is an outline. :) Let me know if you have any edits or have any questions.
Last Name 1
Name
Instructor's name
Course
Date
Laboratory Report
Part 1: Data and Calculations Table 1 Hooke’s Law
Is 𝑦0 = 0? If not 𝑦0 = 15.3 𝑐𝑚 = 0.153 𝑚
Mass m (g)
m (kg)
F=mg (N)
y (cm)
y (m)
50
0.05
0.4905
28.3
0.283
70
0.07
0.6867
34.7
0.347
90
0.09
0.8829
40.8
0.408
110
0.11
1.0791
46.5
0.465
130
0.13
1.2753
52.8
0.528
i)
The graph of force against displacement is as shown below:
Last Name 2
A Graph of Force Against Dispacement
1.4
∆𝐹
1.2
0.8−0.44
𝑠𝑙𝑜𝑝𝑒 = ∆𝑦 = 0.36−0.2 =2.25 N/m
Force, F (N)
1
0.8
∆𝐹
0.6
0.4
∆𝑦
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
Displacement, y(m)
ii)
The best line of fit is drawn and its slope calculated as shown:
𝑆𝑙𝑜𝑝𝑒 = 2.25 𝑁/𝑚
Thus, 𝑆𝑙𝑜𝑝𝑒 = 𝐾 = 2.25 𝑁/𝑚. This is because the slope from the graph of F vs y gives the
constant k as p...