A string on an instrument is tuned to vibrate at a fundamental frequency of 67 Hz. If the tension in the string were increased by a factor of four, what would be the new fundamental frequency?

The new fundamental frequency is doubled! That is, new fundamental frequency = 134 Hz

Explanation:

The frequency (f) of vibration of stringed instruments is related to the Tension (T) in the spring by the relation

fₙ = (n/2L)√(T/μ)

where n = 1,2,3,4...

For fundamental frequency, n = 1

L = length of the string

T = tension in the string

μ = linear density

So, if all the other parameters are constant,

fundamental frequency is proportional to the square root of tension in the string

f ∝√T

f = k√T

k = constant of proportionality

k = f/√T

Let the initial fundamental frequency and tension be f₁ and T₁ respectively

f₁ = 67 Hz, T₁ = ?

k = 67/√T₁

Let the new fundamental frequency and tension be be f₂ and T₂ respectively

f₂ = ? But, T₂ = 4T₁

f₂ = k√T₂ = k√(4T₁) = 2k√T₁

But, k = 67/√T₁

f₂ = 2× (67/√T₁) × √T₁ = 2 × 67 = 134 Hz

Hence the new fundamental frequency is doubled!.

Explanation:

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terd muffin is the correct answer