Question
3. If the 3rd harmonic has a frequency of 415 Hz and the wire has a length of 120 cm, determine the number of nodes , antinodes , standing waves , wavelength, and the speed of the wave.
Solution
4.1
(118 Votes)
Preston
Professional · Tutor for 6 years
Answer
Here's the solution:**1. Understanding Harmonics:**The 3rd harmonic means the string is vibrating in a pattern with three loops. A harmonic is a multiple of the fundamental frequency (the lowest frequency at which the string can vibrate).**2. Nodes and Antinodes:*** **Nodes:** Points on the string that do not move (zero displacement).* **Antinodes:** Points on the string with maximum displacement.For the 3rd harmonic:* **Nodes:** There are 4 nodes (including the two fixed ends).* **Antinodes:** There are 3 antinodes.**3. Standing Waves:**The 3rd harmonic itself represents a standing wave. So, there's one standing wave present, specifically the 3rd harmonic standing wave.**4. Wavelength (λ):**In the 3rd harmonic, three half-wavelengths fit within the length (L) of the string:3(λ/2) = Lλ = 2L/3λ = (2 * 120 cm) / 3λ = 80 cm = 0.8 m**5. Speed of the Wave (v):**We can use the wave equation:v = fλwhere:* v is the speed of the wave* f is the frequency* λ is the wavelengthv = 415 Hz * 0.8 mv = 332 m/s**Summary of Answers:*** **Nodes:** 4* **Antinodes:** 3* **Standing Waves:** 1 (the 3rd harmonic)* **Wavelength:** 80 cm or 0.8 m* **Speed of the wave:** 332 m/s