The frequency of vibration of a string depends on the length l

  1. 16.4: Wave Speed on a Stretched String
  2. The Frequency of Vibration of a String Depends on the Length L Between the Nodes, the Tension F in the String and Its Mass per Unit Length M. Guess the Expression for
  3. The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m, Guess the expression for its frequency from dimensional analysis.
  4. The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length mu of string. Derive the formula for frequency.
  5. String vibration
  6. The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length mu of string. Derive the formula for frequency.
  7. 16.4: Wave Speed on a Stretched String
  8. String vibration
  9. The Frequency of Vibration of a String Depends on the Length L Between the Nodes, the Tension F in the String and Its Mass per Unit Length M. Guess the Expression for
  10. The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m, Guess the expression for its frequency from dimensional analysis.


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16.4: Wave Speed on a Stretched String

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F16%253A_Waves%2F16.04%253A_Wave_Speed_on_a_Stretched_String \( \newcommand\) • • • • • • • Learning Objectives • Determine the factors that affect the speed of a wave on a string • Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media The speed of a wave depends on the characteristics of the medium. For example, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness but may be made of similar material. They have different linear densities, where the linear density is defined as the mass per length, \[\mu = \frac\, kg. \nonumber\] The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed. Wave Speed on a String under Tension To see how the speed of a wave on a string depends on ...

The Frequency of Vibration of a String Depends on the Length L Between the Nodes, the Tension F in the String and Its Mass per Unit Length M. Guess the Expression for

Frequency,f \[\propto\] L aF bm c f=kL aF bm c ...(1) Dimension of [f] = [T −1] Dimension of the right side components: [L] = [L] [F] = [MLT −2] [m] = [ML −1] Writing equation (1) in dimensional form, we get: [T −1] = [L] a[MLT −2] b[ML −1] c [M 0L 0T −1] = [M b+cL a+b−cT −2b] Equating the dimensions of both sides, we get: b+c= 0 ....(i) a+b−c= 0 ....(ii) −2b = −1 ....(iii) Solving equations (i), (ii) and (iii), we get:

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m, Guess the expression for its frequency from dimensional analysis.

Expression of frequency is given as follows: Frequency, f ∝ [ L a ] . . . ( 1 ) f ∝ [ F b ] . . . ( 2 ) f ∝ [ M c ] . . . ( 3 ) Combining eAq(1) (2) and (3) we can say: f = [ K L a F b M c ] where M = Mass / unit length L = Length F = Tension (Force) Dimension of . f = [ T − 1 ] Dimension of right side: Dimension of force, F = [ M L T − 2 ] b = [ M b L b T − 2 b ] Dimension of mass per unit length, = [ M L − 1 ] c = [ M c L − c ] So, [ T − 1 ] = [ L a ] [ M b L b T − 2 b ] [ M c L − c ] [ M 0 L 0 T − 1 ] = = M b + c L a + b − c T − 2 b Equating the dimensions of both sides, we get b +c = 0 b +c = 0 ......(i) - c + a + b = 0 - c + a + b = 0 ......(ii) - 2 b = - 1 - 2 b = - 1 ......(iii) Solving the equations, we get, \ a = − 1 , b = 1 2 and c = − 1 2 ∴ f = K L − 1 F 1 2 M − 1 2 Hence expression of frequency will be as follows: f = K L √ F M

The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length mu of string. Derive the formula for frequency.

Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b and mass per unit length raised to the power c. Then, f ∝ [ F ] a [ l ] b [ μ ] c or, f = k [ F ] a [ l ] b [ μ ] c ...(i) Here, k is a dimensionless constant. Thus, [ f ] = [ F ] a [ l ] b [ μ ] c or, [ M 0 L 0 T − 1 ] = [ M L T − 2 ] e [ L ] b [ M L − 1 ] c or, [ M 0 L 0 T − 1 ] = [ M a + c L a + b − c T − 2 a ] For dimensional balance, the dimensions on both sides should be same. Thus, a + c = 0 ...(ii) a + b − c = 0 ...(iii) − 2 a = − 1 ...(iv) Solving these three equations, we get a = 2 1 ​ , c = − 2 1 ​ a n d b = − 1 Substituting these values in Eq. (i), we get f = k ( F ) 1 / 2 ( l ) − 1 ( μ ) − 1 / 2 o r f = l k ​ μ F ​ ​ Experimentally, the value of k is found to be 2 1 ​ . Hence, f = 2 l 1 ​ μ F ​ ​

String vibration

A vibrating string to produce a Wave [ ] The velocity of propagation of a wave in a string ( v This relationship was discovered by [ citation needed] Derivation [ ] Source: Let Δ x Observing string vibrations [ ] One can see the not of an analog oscilloscope). This effect is called the A similar but more controllable effect can be obtained using a Real-world example [ ] See also: A Wikipedia user's L Repeating this computation for all six strings results in the following frequencies. Shown next to each frequency is the musical note (in Fundamental harmonics as computed by above string vibration formulas String no. Computed frequency [Hz] Closest note in 1 330 E 4 (= 440 ÷ 2 5/12 ≈ 329.628 Hz) 2 247 B 3 (= 440 ÷ 2 10/12 ≈ 246.942 Hz) 3 196 G 3 (= 440 ÷ 2 14/12 ≈ 195.998 Hz) 4 147 D 3 (= 440 ÷ 2 19/12 ≈ 146.832 Hz) 5 110 A 2 (= 440 ÷ 2 24/12 = 110 Hz) 6 82.4 E 2 (= 440 ÷ 2 29/12 ≈ 82.407 Hz) See also [ ] • • • • • • • References [ ] • Molteno, T. C. A.; N. B. Tufillaro (September 2004). "An experimental investigation into the dynamics of a string". American Journal of Physics. 72 (9): 1157–1169. • Tufillaro, N. B. (1989). "Nonlinear and chaotic string vibrations". American Journal of Physics. 57 (5): 408. Specific

The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length mu of string. Derive the formula for frequency.

Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b and mass per unit length raised to the power c. Then, f ∝ [ F ] a [ l ] b [ μ ] c or, f = k [ F ] a [ l ] b [ μ ] c ...(i) Here, k is a dimensionless constant. Thus, [ f ] = [ F ] a [ l ] b [ μ ] c or, [ M 0 L 0 T − 1 ] = [ M L T − 2 ] e [ L ] b [ M L − 1 ] c or, [ M 0 L 0 T − 1 ] = [ M a + c L a + b − c T − 2 a ] For dimensional balance, the dimensions on both sides should be same. Thus, a + c = 0 ...(ii) a + b − c = 0 ...(iii) − 2 a = − 1 ...(iv) Solving these three equations, we get a = 2 1 ​ , c = − 2 1 ​ a n d b = − 1 Substituting these values in Eq. (i), we get f = k ( F ) 1 / 2 ( l ) − 1 ( μ ) − 1 / 2 o r f = l k ​ μ F ​ ​ Experimentally, the value of k is found to be 2 1 ​ . Hence, f = 2 l 1 ​ μ F ​ ​

16.4: Wave Speed on a Stretched String

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F16%253A_Waves%2F16.04%253A_Wave_Speed_on_a_Stretched_String \( \newcommand\) • • • • • • • Learning Objectives • Determine the factors that affect the speed of a wave on a string • Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media The speed of a wave depends on the characteristics of the medium. For example, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness but may be made of similar material. They have different linear densities, where the linear density is defined as the mass per length, \[\mu = \frac\, kg. \nonumber\] The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed. Wave Speed on a String under Tension To see how the speed of a wave on a string depends on ...

String vibration

A vibrating string to produce a Wave [ ] The velocity of propagation of a wave in a string ( v This relationship was discovered by [ citation needed] Derivation [ ] Source: Let Δ x Observing string vibrations [ ] One can see the not of an analog oscilloscope). This effect is called the A similar but more controllable effect can be obtained using a Real-world example [ ] See also: A Wikipedia user's L Repeating this computation for all six strings results in the following frequencies. Shown next to each frequency is the musical note (in Fundamental harmonics as computed by above string vibration formulas String no. Computed frequency [Hz] Closest note in 1 330 E 4 (= 440 ÷ 2 5/12 ≈ 329.628 Hz) 2 247 B 3 (= 440 ÷ 2 10/12 ≈ 246.942 Hz) 3 196 G 3 (= 440 ÷ 2 14/12 ≈ 195.998 Hz) 4 147 D 3 (= 440 ÷ 2 19/12 ≈ 146.832 Hz) 5 110 A 2 (= 440 ÷ 2 24/12 = 110 Hz) 6 82.4 E 2 (= 440 ÷ 2 29/12 ≈ 82.407 Hz) See also [ ] • • • • • • • References [ ] • Molteno, T. C. A.; N. B. Tufillaro (September 2004). "An experimental investigation into the dynamics of a string". American Journal of Physics. 72 (9): 1157–1169. • Tufillaro, N. B. (1989). "Nonlinear and chaotic string vibrations". American Journal of Physics. 57 (5): 408. Specific

The Frequency of Vibration of a String Depends on the Length L Between the Nodes, the Tension F in the String and Its Mass per Unit Length M. Guess the Expression for

Frequency,f \[\propto\] L aF bm c f=kL aF bm c ...(1) Dimension of [f] = [T −1] Dimension of the right side components: [L] = [L] [F] = [MLT −2] [m] = [ML −1] Writing equation (1) in dimensional form, we get: [T −1] = [L] a[MLT −2] b[ML −1] c [M 0L 0T −1] = [M b+cL a+b−cT −2b] Equating the dimensions of both sides, we get: b+c= 0 ....(i) a+b−c= 0 ....(ii) −2b = −1 ....(iii) Solving equations (i), (ii) and (iii), we get:

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m, Guess the expression for its frequency from dimensional analysis.

Expression of frequency is given as follows: Frequency, f ∝ [ L a ] . . . ( 1 ) f ∝ [ F b ] . . . ( 2 ) f ∝ [ M c ] . . . ( 3 ) Combining eAq(1) (2) and (3) we can say: f = [ K L a F b M c ] where M = Mass / unit length L = Length F = Tension (Force) Dimension of . f = [ T − 1 ] Dimension of right side: Dimension of force, F = [ M L T − 2 ] b = [ M b L b T − 2 b ] Dimension of mass per unit length, = [ M L − 1 ] c = [ M c L − c ] So, [ T − 1 ] = [ L a ] [ M b L b T − 2 b ] [ M c L − c ] [ M 0 L 0 T − 1 ] = = M b + c L a + b − c T − 2 b Equating the dimensions of both sides, we get b +c = 0 b +c = 0 ......(i) - c + a + b = 0 - c + a + b = 0 ......(ii) - 2 b = - 1 - 2 b = - 1 ......(iii) Solving the equations, we get, \ a = − 1 , b = 1 2 and c = − 1 2 ∴ f = K L − 1 F 1 2 M − 1 2 Hence expression of frequency will be as follows: f = K L √ F M

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