Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

No Downloads

Total views

0

On SlideShare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

63

Comments

0

Likes

4

No embeds

No notes for slide

- 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

No public clipboards found for this slide

Be the first to comment