Chapter 15, Problem 33P

(a)

To determine

To Find:The derivative of wave speed on a string with respect to the tension.

Explanation of Solution

Given:

Differentials $\text{dv}$ and ${\text{dF}}_{\text{T}}$ obey $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{dF}}_{\text{T}}{\text{/F}}_{\text{T}}$ .

Formula used:

The speed of the transverse wave is given by $\text{v=}\sqrt{{\text{F}}_{\text{T}}\text{/μ}}$ .

Where,

  ${\text{F}}_{\text{T}}\text{=}$ Tension in the wire

  $\text{μ=}$ Linear density

Calculation:

For differentiating the expression with respect to ${\text{F}}_{\text{T}}$ and then separatethe variables to show that the differentials satisfy $\text{dv/v=}\frac{\text{1}}{\text{2}}{\text{dF}}_{\text{T}}{\text{/F}}_{\text{T}}$ .

For evaluating the ${\text{dv/dF}}_{\text{T}}$ is

  $\begin{array}{l}\frac{\text{dv}}{\text{dF}}\text{=}\frac{\text{d}}{{\text{dF}}_{\text{T}}}\left[\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}\right]\\ \text{=}\frac{\text{1}}{\text{2}}\sqrt{\frac{\text{1}}{{\text{F}}_{\text{T}}\text{μ}}}\\ \text{=}\frac{\text{1}}{\text{2}}\frac{\text{v}}{{\text{F}}_{\text{T}}}\end{array}$

Now, to separate the variables to obtain

  $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{dF}}_{\text{T}}}{{\text{F}}_{\text{T}}}$

Conclusion:

Thus, derivative of the speed of the wave on a string with respect to the tension is $\frac{\text{1}}{\text{2}}\frac{{\text{dF}}_{\text{T}}}{{\text{F}}_{\text{T}}}$ .

(b)

To determine

To Calculate:The tension that must be changed to increase the speed to $\text{312}\text{m/s}$ using differential approximation.

Answer to Problem 33P

The tension that must be changed to increase the speed to $\text{312}\text{m/s}$ using differential approximation is $\text{40 N}$ .

Explanation of Solution

Given:

Speed of the wave $\text{=300}\text{m/s}$ .

Tension $\text{=500}\text{N}$

The speed is increased to 312 m/s.

Formula used:

  ${\text{dF}}_{\text{T}}$ can be evaluated using:

  ${\text{dF}}_{\text{T}}{\text{=2F}}_{\text{T}}\frac{\text{dv}}{\text{v}}$

Calculation:

To estimate how much tension must be changed to increase the speed of the wave to $\text{312}\text{m/s}$ , substitute the values:

Approximate the ${\text{dF}}_{\text{T}}$ with ${\text{ΔF}}_{\text{T}}$ and $\text{dv}$ with $\text{Δv}$ to get

  ${\text{ΔF}}_{\text{T}}{\text{=2F}}_{\text{T}}\frac{\text{Δv}}{\text{v}}$

Put the numerical values to get ${\text{ΔF}}_{\text{T}}$ ,

  $\begin{array}{l}{\text{ΔF}}_{\text{T}}{\text{=2F}}_{\text{T}}\frac{\text{Δv}}{\text{v}}\\ {\text{ΔF}}_{\text{T}}\text{=2}\left(\text{500}\text{N}\right)\left(\frac{\text{312}\text{m/s-300}\text{m/s}}{\text{300}\text{m/s}}\right)\\ {\text{ΔF}}_{\text{T}}\text{=40 N}\end{array}$

Conclusion:

Thus, the tension that must be changed to increase the speed to $\text{312}\text{m/s}$ using differential approximation is $\text{40 N}$ .

(c)

To determine

To Calculate: ${\text{ΔF}}_{\text{T}}$ and compare it to the differential result in part (b).

Answer to Problem 33P

The value of the ${\text{ΔF}}_{\text{T}}$ is $\text{40}\text{.8}\text{N}$ and after ${\text{ΔF}}_{\text{T}}$ comparing it to the differential result in part (b) percent error is $\text{2}\text{\%}$ .

Explanation of Solution

Given:

Speed of the wave $\text{=300}\text{m/s}$ .

Tension $\text{=500}\text{N}$

Formula used:

Wave speed of a transverse wave is given by:

  $\text{v=}\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}$

Calculation:

The exact value for ${\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}$ is given by

  ${\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{=F}}_{\text{T,2}}{\text{-F}}_{\text{T,1}}\mathrm{.....}\left(\text{1}\right)$

Express the wave speeds for the two tensions

  $\begin{array}{l}{\text{v}}_{\text{1}}\text{=}\sqrt{\frac{{\text{F}}_{\text{T,1}}}{\text{μ}}}\\ {\text{v}}_{\text{2}}\text{=}\sqrt{\frac{{\text{F}}_{\text{T,1}}}{\text{μ}}}\end{array}$

After that dividing the second equation by the first one, it simply yields:

  $\begin{array}{l}\frac{{\text{v}}_{\text{2}}}{{\text{v}}_{\text{1}}}\text{=}\frac{\sqrt{\frac{{\text{F}}_{\text{T,2}}}{\text{μ}}}}{\sqrt{\frac{{\text{F}}_{\text{T,1}}}{\text{μ}}}}\\ {\text{v}}_{\text{2}}\text{=}\sqrt{\frac{{\text{F}}_{\text{T,2}}}{{\text{F}}_{\text{T,1}}}}\\ {\text{F}}_{\text{T,2}}{\text{=F}}_{\text{T,1}}{\left(\frac{{\text{v}}_{\text{2}}}{{\text{v}}_{\text{1}}}\right)}^{\text{2}}\end{array}$

Put the value of ${\text{F}}_{\text{T,2}}$ in first equation yields,

  $\begin{array}{l}{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{=F}}_{\text{T,2}}{\text{-F}}_{\text{T,1}}\\ {\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{=F}}_{\text{T,1}}\left[{\left(\frac{{\text{v}}_{\text{2}}}{{\text{v}}_{\text{1}}}\right)}^{\text{2}}\text{-1}\right]\end{array}$

Put the numerical values to evaluate ${\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}$ ,

  $\begin{array}{l}{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{=F}}_{\text{T,2}}{\text{-F}}_{\text{T,1}}\\ {\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{=F}}_{\text{T,1}}\left[{\left(\frac{{\text{v}}_{\text{2}}}{{\text{v}}_{\text{1}}}\right)}^{\text{2}}\text{-1}\right]\\ {\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}\text{=}\left(\text{500}\text{N}\right)\left[{\left(\frac{\text{312}\text{m/s}}{\text{300}\text{m/s}}\right)}^{\text{2}}\text{-1}\right]\\ {\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}\text{=40}\text{.8}\text{N}\end{array}$

Now, to find the percent error between the exact and approximate values for ${\text{ΔF}}_{\text{T}}$ is,

  $\begin{array}{l}\frac{{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{-ΔF}}_{\text{T}}}{{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}}\text{=}\frac{\text{40}\text{.8}\text{N-40}\text{.0}\text{N}}{\text{40}\text{.8}\text{N}}\\ \frac{{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}{\text{-ΔF}}_{\text{T}}}{{\left({\text{ΔF}}_{\text{T}}\right)}_{\text{exact}}}\text{=2}\text{\%}\end{array}$

Conclusion:

Thus, the value of the ${\text{ΔF}}_{\text{T}}$ is $\text{40}\text{.8}\text{N}$ and after ${\text{ΔF}}_{\text{T}}$ comparing it to the differential result in part (b) percent error is $\text{2}\text{\%}$ .