Chapter 15, Problem 102P

(a)

To determine

To calculate: The period and frequency of the waves on string.

Answer to Problem 102P

The period $\left(\text{T}\right)$ is $\text{2}\text{.50}\text{ms}$ and frequency $\left(\text{f}\right)$ is $\text{400}\text{Hz}$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension = $\text{1}\text{.0}\text{KN}$ .

Formula used:

Theperiod and frequency of the waves on string can be calculated as:

  $\text{T=}\frac{\text{1}}{\text{f}}$

Where,

  $\text{T=}$ period.

  $\text{f=}$ frequency of the string.

Calculation:

The frequency of the waves on the string is the similar as that of frequency of the tuning fork and their period is the reciprocal of the frequency.

The frequency of the wave given is:

  $\text{f=400}\text{Hz}$

The period of the wave on the wire is the reciprocal of their frequency:

  $\begin{array}{l}\text{T=}\frac{\text{1}}{\text{f}}\\ \text{T=}\frac{\text{1}}{\text{400}{\text{s}}^{\text{-1}}}\\ \text{T=2}\text{.50}\text{ms}\end{array}$

Conclusion:

Thus, the period $\left(\text{T}\right)$ is $\text{2}\text{.50}\text{ms}$ and frequency $\left(\text{f}\right)$ is $\text{400}\text{Hz}$ .

(b)

To determine

To calculate: The speed of the wave.

Answer to Problem 102P

The speed of the wave $\left(\text{v}\right)$ is $\text{316}\text{.2}\text{m/s}$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension = $\text{1}\text{.0}\text{KN}$ .

Formula used:

For wave speed formula used is:

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

Where,

  $\text{v=}$ velocity.

  ${\text{F}}_{\text{T}}=$ tension in the string.

  $\text{μ=}$ linear density.

Calculation:

By using the tension and the linear density, wave speed can be calculated.

Relate the speed of the waves to the tension in string and linear density:

  $\begin{array}{l}\text{v=}\sqrt{\frac{{\text{F}}_{\text{T}}}{\text{μ}}}\\ \text{v=}\sqrt{\frac{\text{1}\text{.0}\text{kN}}{\text{0}\text{.010}\text{kg/m}}}\\ \text{v=316}\text{.2}\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the wave $\left(\text{v}\right)$ is $\text{316}\text{.2}\text{m/s}$ .

(c)

To determine

To calculate: The wavelength and wave number.

Answer to Problem 102P

The wavelength $\left(\text{λ}\right)$ is $\text{79}\text{cm}$ and $\left(\text{k}\right)$ wave number is $\text{7}\text{.9}{\text{m}}^{\text{-1}}$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension = $\text{1}\text{.0}\text{KN}$ .

Formula used:

  $\text{v= f λ}$

Where,

Sound’s speed: $\text{v}$ .

Frequency of wave: $\text{f }$ .

The wavelength: $\text{λ}$ .

Calculation:

By using the frequency and the speed of the waves and the wave number The wavelength can be determined.

Relate the wavelength and wave no to the speed and frequency of the wave:

  $\text{v= f λ}$

Where,

  $\text{v}$ = Sound’s speed:.

  $\text{f }$ = Frequency of wave.

  $\text{λ}$ = The wavelength.

After substituting the values,

  $\begin{array}{l}\text{v= f λ}\\ \text{λ=}\frac{\text{v}}{\text{f}}\\ \text{λ=}\frac{\text{316}\text{.2}\text{m/s}}{\text{400}{\text{s}}^{\text{-1}}}\\ \text{λ=79}\text{.05}\text{cm}\\ \text{λ=79}\text{cm}\end{array}$

Therefore, $\text{λ=79}\text{cm}$ .

Now, evaluate the wave number using wave length:

  $\begin{array}{l}\text{k=}\frac{\text{2π}}{\text{79}{\text{.05×10}}^{\text{-2}}\text{m}}\\ \text{k=7}\text{.95}{\text{m}}^{\text{-1}}\\ \text{k=7}\text{.9}{\text{m}}^{\text{-1}}\end{array}$

Hence, $\text{k=7}\text{.9}{\text{m}}^{\text{-1}}$ .

Conclusion:

Thus, the wavelength $\left(\text{λ}\right)$ is $\text{79}\text{cm}$ and $\left(\text{k}\right)$ wave number is $\text{7}\text{.9}{\text{m}}^{\text{-1}}$ .

(d)

To determine

To calculate: Suitable wave function for the wave on the string.

Answer to Problem 102P

The suitable wave function is $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{0}\text{.50}\text{mm}\right)\left[\left(\text{7}\text{.9}{\text{m}}^{\text{-1}}\right)\text{x-}\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)\text{t}\right]$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension = $\text{1}\text{.0}\text{KN}$ .

Formula used:

For wave speed formula used is:

  $\text{ω=2πf}$

Where,

  $\text{ω=}$ angular velocity.

Calculation:

The general form of the wave function for waves on a string is $\text{y}\left(\text{x,t}\right)\text{=Asin}\left(\text{kx±ωt}\right)$ .

So, with the help of $\text{k},\text{ω,A}$ wave function is calculated.

Initially, find out the angular frequency of the waves:

  $\begin{array}{l}\text{ω=2πf}\\ \text{ω=2π}\left(\text{400}{\text{s}}^{\text{-1}}\right)\\ \text{ω=2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\end{array}$

Now, put $\text{k},\text{ω,A}$ in general form of the wave function to get:

  $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{0}\text{.50}\text{mm}\right)\left[\left(\text{7}\text{.9}{\text{m}}^{\text{-1}}\right)\text{x-}\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)\text{t}\right]$

Conclusion:

Thus, the suitable wave function is $\text{y}\left(\text{x,t}\right)\text{=}\left(\text{0}\text{.50}\text{mm}\right)\left[\left(\text{7}\text{.9}{\text{m}}^{\text{-1}}\right)\text{x-}\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)\text{t}\right]$ .

(e)

To determine

To calculate: max speed and acceleration point on the string.

Answer to Problem 102P

The max speed $\left({\text{v}}_{\text{max}}\right)$ is $\text{1}\text{.3}\text{m/s}$ and acceleration $\left({\text{a}}_{\text{max}}\right)$ point on the string is $\text{3}\text{.2}{\text{km/s}}^{\text{2}}$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension = $\text{1}\text{.0}\text{KN}$ .

Formula used:

For max speed formula used is:

  ${\text{v}}_{\text{max}}\text{=Aω}$

Where,

  ${\text{v}}_{\text{max}}\text{=}$ max velocity.

  $\text{A=}$ amplitude of the velocity.

  $\text{ω=}$ angular frequency.

Calculation:

The max speed and acceleration ofa point on the string can be determined from the angular frequency and amplitude ofthe waves.

Relate the max speed of apoint on the string to the amplitude of the waves and tuning fork’s the angular frequency:

  $\begin{array}{l}{\text{v}}_{\text{max}}\text{=Aω}\\ {\text{v}}_{\text{max}}\text{=}\left(\text{0}{\text{.50×10}}^{\text{-3}}\text{m}\right)\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)\\ {\text{v}}_{\text{max}}\text{=1}\text{.3}\text{m/s}\end{array}$

Now, expression for the max acceleration of string point in terms of the amplitude and angular frequency of the tuning fork is:

  ${\text{a}}_{\text{max}}{\text{=Aω}}^{\text{2}}$

Put the values to get max acceleration:

  $\begin{array}{l}{\text{a}}_{\text{max}}{\text{=Aω}}^{\text{2}}\\ {\text{a}}_{\text{max}}\text{=}\left(\text{0}{\text{.50×10}}^{\text{-3}}\text{m}\right){\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)}^{\text{2}}\\ {\text{a}}_{\text{max}}\text{=3}\text{.2}{\text{km/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the max speed $\left({\text{v}}_{\text{max}}\right)$ is $\text{1}\text{.3}\text{m/s}$ and acceleration $\left({\text{a}}_{\text{max}}\right)$ point on the string is $\text{3}\text{.2}{\text{km/s}}^{\text{2}}$ .

(f)

To determine

To calculate: minimum average rate of energy supplied to fork.

Answer to Problem 102P

The minimum average rate of energy $\left({\text{p}}_{\text{av}}\right)$ supplied to fork is $\text{2}\text{.5}\text{W}$ .

Explanation of Solution

Given:

Frequency = $\text{400}\text{Hz}$ .

Amplitude = $\text{0}\text{.50}\text{mm}$ .

Linear mass density = $\text{0}\text{.010}\text{kg/m}$ .

Tension: $\text{1}\text{.0}\text{KN}$ .

Formula used:

For minimum average rate of energyformula used is:

  ${\text{p}}_{\text{av}}\text{=}\frac{\text{1}}{\text{2}}{\text{μA}}^{\text{2}}{\text{ω}}^{\text{2}}\text{v}$

Where,

  ${\text{P}}_{\text{av}}\text{=}$ average energy.

  $\text{A=}$ amplitude of the velocity.

  $\text{ω=}$ angular frequency.

  $\text{v=}$ velocity.

Calculation:

The expression for the minimum average power essential to keep the tuning fork oscillating at steady amplitude in terms of linear density of string, the amplitude of its vibrations and wave speed:

  ${\text{p}}_{\text{av}}\text{=}\frac{\text{1}}{\text{2}}{\text{μA}}^{\text{2}}{\text{ω}}^{\text{2}}\text{v}$

Where,

  ${\text{P}}_{\text{av}}\text{=}$ average energy.

  $\text{A=}$ amplitude of the velocity.

  $\text{ω=}$ angular frequency.

  $\text{v=}$ velocity.

Now, substitute the values in the equation:

  $\begin{array}{l}{\text{p}}_{\text{av}}\text{=}\frac{\text{1}}{\text{2}}\left(\text{0}\text{.010}\text{kg/m}\right){\left(\text{2}{\text{.51×10}}^{\text{3}}{\text{s}}^{\text{-1}}\right)}^{\text{2}}\left(\text{0}{\text{.50×10}}^{\text{-3}}\text{m}\right)\left(\text{316}\text{m/s}\right)\\ {\text{p}}_{\text{av}}\text{=2}\text{.5}\text{W}\end{array}$

Conclusion:

Thus, the minimum average rate of energy $\left({\text{p}}_{\text{av}}\right)$ supplied to fork is $\text{2}\text{.5}\text{W}$ .