A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is placed at a distance ℓ from one end, the remaining length L - ℓ of the string vibrates in its first harmonic with a frequency of 463 Hz. What is the distance ℓ? Express your answer as a ratio of the length L.

Question

Physics

Addyson

18 November, 02:27

A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is placed at a distance ℓ from one end, the remaining length L - ℓ of the string vibrates in its first harmonic with a frequency of 463 Hz. What is the distance ℓ? Express your answer as a ratio of the length L.

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Answers (1)

Know the Answer?

Woodard · Answer 1

i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

2 L = n λ n = 1, 2, 3 ...

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

v = λ f

λ = v / f

Let's replace in the first equation

2 L = 1 (v / f₁)

For the shortest length L = L-l

2 (L - l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

v = 2L f₁

v = 2 (L-l) f₂

2L f₁ = 2 (L-l) f₂

L - l = L f₁ / f₂

l = L - L f₁ / f₂

l = L (1 - f₁ / f₂)

.

Let's calculate

l / L = (1 - 309/463)

i / L = 0.3326