Chapter 11, Problem 79GP

To determine

To Find: The maximum speed of the tip of a prong of a tuning fork x.

Answer to Problem 79GP

  $3\text{m/s}$

Explanation of Solution

Given Information:

Frequency of tuning fork, $\mathrm{f}=264\text{Hz}$ .

Amplitude, $\mathrm{A}=18\min =1.8\times {10}^{-3}\text{m}$

Formula used:

Write the expression for the max speed of the tip of prong is

  ${\mathrm{v}}_{\max }=2\mathrm{\pi }\mathrm{f}\mathrm{A}$

Here, $\mathrm{f}$ the frequency.

  $\mathrm{A}$ is the amplitude.

Calculation:

From the above equation

  ${\mathrm{v}}_{\max }=2\mathrm{\pi }(264\text{Hz})(1.8\times {10}^{-3}\text{m})$

  ${\mathrm{v}}_{\max }=3\text{m/s}$

Conclusion: The maximum speed of the tip of a prong of a tuning fork is $3\text{m/s}$ .

To determine

To Find: The maximum acceleration of the tip of a prong.

Answer to Problem 79GP

  $4.9\times {10}^3{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given Information:

Frequency of the tuning fork, $\mathrm{f}=264\text{Hz}$

Formula used:

The acceleration is ${\mathrm{a}}_{\max }={\mathrm{\omega }}^2\times \mathrm{A}$

Here, $\mathrm{\omega }$ is the angular velocity,

  $\mathrm{\omega }=2\mathrm{\pi }\mathrm{f}$

  ${\mathrm{a}}_{\max }={(2\mathrm{\pi }\mathrm{f})}^2\times \mathrm{A}$

Calculation:

Substitute the values in the above expression

  ${\mathrm{a}}_{\max }={\left(2\mathrm{\pi }\times 264\text{Hz}\right)}^2(1.8\times {10}^{-3}\text{m})$

  ${\mathrm{a}}_{\max }=4.9\times {10}^3{\text{m/s}}^{\text{2}}$

Conclusion: The maximum acceleration of the tip of a prong is $4.9\times {10}^3{\text{m/s}}^{\text{2}}$ .