Everything Science Grade 11

16.2 CHAPTER 16. THE PHYSICS OF MUSIC

2. Use the chart to finda formula for the wavelength in terms of the number of nodes.

You should have foundthis formula:

λ =

2 L

n− 1

Here, n is the number of nodes. L is the length of the string. The frequency f is:

f =
v
λ
Here, v is the velocity of the wave. This may seem confusing. The wave is a standing wave, so how
can it have a velocity? But one standing wave ismade up of many wavesthat travel back and forth on
the string. Each of thesewaves has the same velocity. This speed depends on the mass and tension of
the string.

Example 1: Harmonics on a String

QUESTION

We have a standing wave on a string that is 65 cm long. The wave has avelocity of 143

m.s−^1. Find the frequencies of the fundamental, first,second, and third harmonics.

SOLUTION

Step 1 : Identify what is given and what is asked:

L = 65 cm = 0.65 m

v = 143 m.s−^1

f =?

To find the frequency we will use f =vλ

Step 2 : Find the wavelength for each harmonic:

To find f we need the wavelength of each harmonic (λ =n^2 −L 1 ). The

wavelength is then substituted into f =vλto find the harmonics. The table

below shows the calculations.

Nodes

Wavelength

λ =n^2 −L 1

Frequency

f =vλ

Fundamental frequency fo 2 2(0 2 −,65) 1 = 1,3^1431 , 3 = 110 Hz

First harmonic f 1 3 2(0 3 −,65) 1 = 0, (^650143) , 65 = 220 Hz
Second harmonic f 2 4 2(0 4 −,65) 1 = 0, (^430143) , 43 = 330 Hz
Third harmonic f 3 5 2(0 5 −,65) 1 = 0, (^330143) , 33 = 440 Hz
110 Hz is the natural frequency of the A string on a guitar. The third
harmonic, at 440 Hz, isthe note that orchestrasuse for tuning.