PHY26041/102/0/2026 iQ Level: 0793226427
UNISA
PHY2606 Assignment 3 full solutions 2026
Assignment 3
Due Date: 22nd June 2026 at 09:00
Assignment Unique number: 209217
, ASSIGNMENT 3
Question 1 — G-string on a guitar
Given: linear mass density 𝜇 = 3g/m = 0.003kg/m, length 𝐿 = 63cm = 0.63m, fundamental
frequency 𝜈1 = 196Hz.
(a) Tension in the tuned string
For a string fixed at both ends, the fundamental frequency is:
1 𝑇
𝜈1 = √
2𝐿 𝜇
Solving for tension 𝑇:
𝑇 = 𝜇(2𝐿𝜈1 )2 = 0.003 × (2 × 0.63 × 196)2
= 0.003 × (246.96)2 = 0.003 × 60,989
𝑇 = 182.97 ≈ 183 N
(b) Wavelengths of first three harmonics
2𝐿
For a string fixed at both ends: 𝜆𝑛 = 𝑛
2 × 0.63
𝜆1 = = 1.26 m
1
2 × 0.63
𝜆2 = = 0.63 m
2
2 × 0.63
𝜆3 = = 0.42 m
3
2
UNISA
PHY2606 Assignment 3 full solutions 2026
Assignment 3
Due Date: 22nd June 2026 at 09:00
Assignment Unique number: 209217
, ASSIGNMENT 3
Question 1 — G-string on a guitar
Given: linear mass density 𝜇 = 3g/m = 0.003kg/m, length 𝐿 = 63cm = 0.63m, fundamental
frequency 𝜈1 = 196Hz.
(a) Tension in the tuned string
For a string fixed at both ends, the fundamental frequency is:
1 𝑇
𝜈1 = √
2𝐿 𝜇
Solving for tension 𝑇:
𝑇 = 𝜇(2𝐿𝜈1 )2 = 0.003 × (2 × 0.63 × 196)2
= 0.003 × (246.96)2 = 0.003 × 60,989
𝑇 = 182.97 ≈ 183 N
(b) Wavelengths of first three harmonics
2𝐿
For a string fixed at both ends: 𝜆𝑛 = 𝑛
2 × 0.63
𝜆1 = = 1.26 m
1
2 × 0.63
𝜆2 = = 0.63 m
2
2 × 0.63
𝜆3 = = 0.42 m
3
2
