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professor190 [17]
3 years ago
11

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 pla

ced 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.
Physics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

 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

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