Answer:
(1)
And we have this other relationship between the linear speed and the angular speed:
(2)
We can find the linear velocity like this:
![v = 0.08 m * 3.49 \frac{rad}{s}= 0.279 m/s](https://tex.z-dn.net/?f=%20v%20%3D%200.08%20m%20%2A%203.49%20%5Cfrac%7Brad%7D%7Bs%7D%3D%200.279%20m%2Fs)
And then from equation (1) we can solve for the frequency and we got:
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%20%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
And replacing we got:
![\lambda = \frac{0.279 m/s}{4000 Hz}= 6.98x10^{-5} m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B0.279%20m%2Fs%7D%7B4000%20Hz%7D%3D%206.98x10%5E%7B-5%7D%20m)
And that represent the wavelength in th groove for this case.
Explanation:
For this case we have the following data given:
represent the angular velocity
represent the frequency
we assume that this respresent the distance from the center
We know the following relationship between the wavelength
and the velocity of a wave:
(1)
And we have this other relationship between the linear speed and the angular speed:
(2)
We can find the linear velocity like this:
![v = 0.08 m * 3.49 \frac{rad}{s}= 0.279 m/s](https://tex.z-dn.net/?f=%20v%20%3D%200.08%20m%20%2A%203.49%20%5Cfrac%7Brad%7D%7Bs%7D%3D%200.279%20m%2Fs)
And then from equation (1) we can solve for the frequency and we got:
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%20%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
And replacing we got:
![\lambda = \frac{0.279 m/s}{4000 Hz}= 6.98x10^{-5} m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B0.279%20m%2Fs%7D%7B4000%20Hz%7D%3D%206.98x10%5E%7B-5%7D%20m)
And that represent the wavelength in th groove for this case.