Consider a 10 000-Hz sound recorded on a phonograph record which rotates at $33 \frac{1}{3}$ rev/min. How far apart are the crests of the wave for this sound on the record

(A) at the outer edge of the record, 6.0 inches from the center?

(B) at the inner edge, 1.0 inch from the center?

Question

Consider a 10 000-Hz sound recorded on a phonograph record which rotates at [latex] 33 \frac{1}{3} [/latex] rev/min. How far apart are the crests of the wave for this sound on the record

(A) at the outer edge of the record, 6.0 inches from the center?

(B) at the inner edge, 1.0 inch from the center?

Accepted Answer

(A) The linear speed v of a point at the outer edge of the record is 2πr/T where T is the period of the rotation and r is the distance from the center. We first find T:

[latex] T=\frac{1}{f}=\frac{1}{33.33 rev / min }=0.030 \min \left(\frac{60 s }{1 min } ight)=1.8 s [/latex].

Now, the linear speed at the outer edge is

[latex] v=\frac{2 \pi r}{T}=\frac{2 \pi(6.0 in .)}{1.8 s }=21 in . / s \left(\frac{2.54 cm }{1 in } ight) [/latex].

= 53 cm/s.

Thus, the wave on the record is moving past the needle at this speed. The wavelength is

[latex] \lambda=\frac{v}{f}=\frac{53 cm / s }{10000 Hz }=5.3 imes 10^{-5} m [/latex].

[latex] =53 \mu m [/latex].

(B) The linear speed at the inner edge is

[latex] v=\frac{2 \pi r}{T}=\frac{2 \pi(1.0 in .)}{1.8 s }=3.5 in . / s \left(\frac{2.54 cm }{1 in .} ight) [/latex].

= 8.9 cm/s.

The wavelength is

[latex] \lambda=\frac{v}{f}=\frac{8.9 cm / s }{10000 Hz }=8.9 imes 10^{-6} m [/latex].

[latex] =8.9 \mu m [/latex].

Thus, the problem with noise interfering with the recorded sound is more severe at the inner edge of the disk than at the outer edge.

Question 17.7: Consider a 10 000-Hz sound recorded on a phonograph record w...

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