Math Problem Statement

The frequency of vibrations of a vibrating string is given by f = 1 2L

T 𝜌 where L is the length of the string, T is its tension, and 𝜌 is its linear density.† (a) Find the rate of change of the frequency with respect to the following. (i) the length (when T and 𝜌 are constant)

Incorrect: Your answer is incorrect. (ii) the tension (when L and 𝜌 are constant)

Incorrect: Your answer is incorrect. (iii) the linear density (when L and T are constant)

Solution

The formula for the frequency of vibrations of a vibrating string is:

f=12LTρf = \frac{1}{2L} \sqrt{\frac{T}{\rho}}

where:

  • ff is the frequency,
  • LL is the length of the string,
  • TT is the tension,
  • ρ\rho is the linear density.

We need to find the rate of change of the frequency ff with respect to:

  1. LL (when TT and ρ\rho are constant),
  2. TT (when LL and ρ\rho are constant),
  3. ρ\rho (when LL and TT are constant).

(i) Rate of change of frequency with respect to LL:

We differentiate ff with respect to LL, treating TT and ρ\rho as constants:

f=12LTρf = \frac{1}{2L} \sqrt{\frac{T}{\rho}}

Differentiate with respect to LL:

dfdL=βˆ’12L2Tρ\frac{df}{dL} = -\frac{1}{2L^2} \sqrt{\frac{T}{\rho}}

So, the rate of change of frequency with respect to LL is:

dfdL=βˆ’fL\frac{df}{dL} = -\frac{f}{L}

(ii) Rate of change of frequency with respect to TT:

Differentiate ff with respect to TT, treating LL and ρ\rho as constants:

f=12LTρf = \frac{1}{2L} \sqrt{\frac{T}{\rho}}

Differentiate with respect to TT:

dfdT=12Lβ‹…12Tρ⋅1ρ\frac{df}{dT} = \frac{1}{2L} \cdot \frac{1}{2\sqrt{T\rho}} \cdot \frac{1}{\rho}

Simplify:

dfdT=14Lβ‹…1Tρ=f2T\frac{df}{dT} = \frac{1}{4L} \cdot \frac{1}{\sqrt{T\rho}} = \frac{f}{2T}

So, the rate of change of frequency with respect to TT is:

dfdT=f2T\frac{df}{dT} = \frac{f}{2T}

(iii) Rate of change of frequency with respect to ρ\rho:

Differentiate ff with respect to ρ\rho, treating LL and TT as constants:

f=12LTρf = \frac{1}{2L} \sqrt{\frac{T}{\rho}}

Differentiate with respect to ρ\rho:

dfdρ=12Lβ‹…βˆ’12Tρ⋅Tρ2\frac{df}{d\rho} = \frac{1}{2L} \cdot \frac{-1}{2\sqrt{T\rho}} \cdot \frac{T}{\rho^2}

Simplify:

dfdρ=βˆ’f2ρ\frac{df}{d\rho} = -\frac{f}{2\rho}

So, the rate of change of frequency with respect to ρ\rho is:

dfdρ=βˆ’f2ρ\frac{df}{d\rho} = -\frac{f}{2\rho}

Summary of results:

  • dfdL=βˆ’fL\frac{df}{dL} = -\frac{f}{L}
  • dfdT=f2T\frac{df}{dT} = \frac{f}{2T}
  • dfdρ=βˆ’f2ρ\frac{df}{d\rho} = -\frac{f}{2\rho}

Let me know if you would like further clarification or details!


5 related questions:

  1. How would the frequency change if the string length is doubled?
  2. What happens to the frequency if the tension is increased by 50%?
  3. How does decreasing the linear density by half affect the frequency?
  4. Can you derive the rate of change of frequency with respect to tension in a similar wave equation?
  5. How would varying both tension and length together affect the frequency?

Tip: When working with partial derivatives, always treat other variables as constants to focus on the change with respect to the variable of interest.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Physics of Vibrations

Formulas

f = (1 / 2L) * sqrt(T / 𝜌)
df/dL = -f / L
df/dT = f / (2T)
df/d𝜌 = -f / (2𝜌)

Theorems

Rate of Change
Chain Rule

Suitable Grade Level

Grades 11-12