Question

A 3.49 rad/s (33 rpm) record has a 3.98-kHz tone cut in the groove. If the groove is located 0.100 m from the center of the record (see drawing), what is the "wavelength" in the groove?

Answer 1

8.77×10⁻⁵ m

Step-by-step explanation:

ω = angular velocity = 3.49 rad/s = 33 rpm

r = Radius of the groove = 0.1 m

f = Frequency = 3.98 kHz

Tangential Velocity of record

v = ωr

⇒v = 3.49×0.1

⇒v = 0.349 m/s

The velocity of the wave is 0.349 m/s

v = fλ

`\lambda=(v)/(f)\\\Rightarrow \lambda=(0.349)/(3980)=8.77* 10^(-5)\ m`

∴ Wavelength in the groove is 8.77×10⁻⁵ m