Question
. A rope has a length of 86cm. If the fundamental frequency is 230 Hz, determine and draw the first 3 harmonics.
Solution
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JoshuaMaster · Tutor for 5 years
Answer
Here's how to determine and draw the first three harmonics of a vibrating rope:<br /><br />**Understanding Harmonics**<br /><br />Harmonics are integer multiples of the fundamental frequency (f<sub>1</sub>). They represent the different standing wave patterns that can be established on the rope.<br /><br />* **First Harmonic (Fundamental Frequency, f<sub>1</sub>):** The simplest standing wave with one antinode (point of maximum displacement) in the middle and two nodes (points of zero displacement) at the ends.<br />* **Second Harmonic (f<sub>2</sub>):** Two antinodes and three nodes. The frequency is twice the fundamental frequency (f<sub>2</sub> = 2 * f<sub>1</sub>).<br />* **Third Harmonic (f<sub>3</sub>):** Three antinodes and four nodes. The frequency is three times the fundamental frequency (f<sub>3</sub> = 3 * f<sub>1</sub>).<br /><br />**Calculations**<br /><br />Given:<br /><br />* Fundamental frequency (f<sub>1</sub>) = 230 Hz<br /><br />We can calculate the frequencies of the other harmonics:<br /><br />* Second Harmonic (f<sub>2</sub>) = 2 * f<sub>1</sub> = 2 * 230 Hz = 460 Hz<br />* Third Harmonic (f<sub>3</sub>) = 3 * f<sub>1</sub> = 3 * 230 Hz = 690 Hz<br /><br />**Drawings**<br /><br />Here's how to draw the first three harmonics:<br /><br />1. **First Harmonic (f<sub>1</sub> = 230 Hz):**<br /><br />```<br /> A<br /> / \<br /> / \<br /> N-----N <br />```<br /> * A = Antinode<br /> * N = Node<br /><br />2. **Second Harmonic (f<sub>2</sub> = 460 Hz):**<br /><br />```<br /> A A<br /> / \ / \<br /> / \ / \<br /> N-----N-----N<br />```<br /><br />3. **Third Harmonic (f<sub>3</sub> = 690 Hz):**<br /><br />```<br /> A A A<br /> / \ / \ / \<br /> / \ / \ / \<br /> N-----N-----N-----N<br />```<br /><br />**Relationship to Rope Length**<br /><br />The wavelength (λ) of the fundamental frequency is related to the length (L) of the rope by: λ<sub>1</sub> = 2L. Since the rope is 86 cm long, the wavelength of the fundamental frequency is 2 * 86 cm = 172 cm. The wavelengths of the higher harmonics are related to the fundamental wavelength: λ<sub>n</sub> = λ<sub>1</sub> / n, where n is the harmonic number. While the rope length is important for determining the fundamental frequency, it's not directly used in drawing the *shape* of the harmonic waves. The drawings focus on the relative positions of nodes and antinodes.<br />
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