Question

A 3.49 rad/s (33 1/3 rpm) record has a 4.00 kHz tone cut in the groove. If the groove is located d = 0.080 m from the center of the record (see drawing), what is the wavelength in the groove?

Answer 1

Answer:

$v = \lambda f$   (1)

And we have this other relationship between the linear speed and the angular speed:

$v = rw$  (2)

We can find the linear velocity like this:

$v = 0.08 m * 3.49 \frac{rad}{s}= 0.279 m/s$

And then from equation (1) we can solve for the frequency and we got:

$\lambda = \frac{v}{f}$

And replacing we got:

$\lambda = \frac{0.279 m/s}{4000 Hz}= 6.98x10^{-5} m$

And that represent the wavelength in th groove for this case.

Explanation:

For this case we have the following data given:

$w = 3.49 rad/s$ represent the angular velocity

$f = 4 kHz *\frac{1000 Hz}{1 kHz}= 4000 Hz$ represent the frequency

$r = 0.08 m$ we assume that this respresent the distance from the center

We know the following relationship between the wavelength $\lambda$ and the velocity of a wave:

$v = \lambda f$   (1)

And we have this other relationship between the linear speed and the angular speed:

$v = rw$  (2)

We can find the linear velocity like this:

$v = 0.08 m * 3.49 \frac{rad}{s}= 0.279 m/s$

And then from equation (1) we can solve for the frequency and we got:

$\lambda = \frac{v}{f}$

And replacing we got:

$\lambda = \frac{0.279 m/s}{4000 Hz}= 6.98x10^{-5} m$

And that represent the wavelength in th groove for this case.