Physics 101 LO 6

1. STANDING WAVES
ON
VIOLIN STRINGS

2. STANDING WAVES
• Are stationary (as opposed to travelling waves)
• Vs
STANDING WAVE TRAVELLING WAVE

3. STANDING WAVES
• Are the superposition of two
harmonic waves with equal
amplitude, frequency and
wavelengths but moving in
opposite direction
v
v
Resulting Standing Wave
from adding the two
harmonic waves

4. STANDING WAVES
• Can be generated by plucking a string with
both ends fixed
• Nodes are points with zero amplitudes
• Antinodes are points with maximum
amplitudes

5. STANDING WAVES ON STRINGS
• Strings with two fixed ends can only produce
standing waves with an integral number of half
wavelength called normal modes
• λ 𝑛 =
2𝐿
𝑛
where L = string length
n = number of antinodes = 1, 2, 3, 4, …
• The fundamental frequency (1st harmonic) is the
lowest frequency (longest wavelength)
• 𝑓1 =
1
2𝐿
𝑇
𝜇
where T = tension in the string
𝜇 = linear mass density of the string =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
• The allowed frequencies are called harmonics
• 𝑓𝑛 = n𝑓1 n = 1, 2, 3, 4, …

6. QUESTION PART 1
Tom wants to make a violin for his sister as a birthday present.
Violins usually make sound frequencies ranging from
200~3000Hz. He has a few 30 cm long strings with linear mass
densities:
A 2.8 × 10−4
kg/m
B 4.0 × 10−4
kg/m
C 0.62 g/m
Which string should he use to make the violin in order to get a
fundamental frequency of 700Hz if the tension in the string is
kept at 70 N?

7. Hints
• What variables are given in the question?
• The fundamental frequency (𝑓1), tension (T), and string length (L)

8. Hints
• What variables are given in the question?
• The fundamental frequency (𝑓1), tension (T), and string length (L)
• Which equation to use when solving for linear mass density?
• 𝑓1 =
1
2𝐿
𝑇
𝜇
where T = tension in the string
𝜇 = linear mass density of the string =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔

9. Solution — Tom should use string B
𝑓1 = 700 Hz T = 70 N L = 30 cm = 0.30 m
𝑓1 =
1
2𝐿
𝑇
𝜇
Solve for 𝜇
𝑓1 × 2𝐿 =
𝑇
𝜇
(𝑓1 × 2𝐿) 2
=
𝑇
𝜇
𝜇 =
𝑇
( 𝑓1×2𝐿)
2 =
70 𝑁
(700 𝐻𝑧×2×0.30𝑚)
2 = 3.97× 10−4 kg/ m
≈ 4.0 × 10−4
kg/m

10. QUESTION PART 2
The violin string broke after a few weeks, but Tom
doesn’t have anymore of the same string. If he
uses a string with linear mass density of 4.7 ×
10−4
kg/m, what should the tension be in the
string in order to produce the same sound
frequency (700 Hz)?

11. Hints
• What variables are given in the question?
• The fundamental frequency (𝑓1), linear mass density (𝜇),
and string length (L)

12. Hints
• What variables are given in the question?
• The fundamental frequency (𝑓1), linear mass density (𝜇),
and string length (L)
• Which equation to use when solving for tension?
• 𝑓1 =
1
2𝐿
𝑇
𝜇
where T = tension in the string
𝜇 = linear mass density of the string =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔

13. Solution
𝑓1 = 700 Hz 𝜇 = 4.7 × 10−4
kg/m L = 30 cm = 0.30 m
𝑓1 =
1
2𝐿
𝑇
𝜇
Solve for T
𝑓1 × 2𝐿 =
𝑇
𝜇
(𝑓1 × 2𝐿) 2
=
𝑇
𝜇
T = (𝑓1 × 2𝐿) 2
× 𝜇 = (700𝐻𝑧 × 2 × 0.30𝑚) 2
× 4.7 × 10−4
kg/m
= 82.9 N
≈ 83 N

14. THANK YOU FOR WATCHING