   Chapter 14, Problem 63P

Chapter
Section
Textbook Problem

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 1.10 × 102 Hz has two strings at this frequency. If one string slips from its normal tension of 6.00 × 102 N to 5.40 × 102 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?

To determine
The beat frequency heard when the hammer strikes the two strings simultaneously.

Explanation

Given Info: First string is hit by a hammer and Standing wave is setup in a string with both ends fixed.

Formula to calculate wave length of a wave produced in a string with both ends fixed is,

λ1=2L

• λ1 is wave length of a wave produced in a string.
• L is the length of the string.

Formula to calculate the speed of the wave in a string is,

v=T1μ

• v is velocity of the wave in a string.
• T1 is the tension in the first string.
• μ is the linear density of the strings.

Formula to calculate the frequency of a wave in terms of speed of the wave and wave length is,

f1=vλ1

• f1 is the frequency of the wave.

Substitute 2L for λ1 , T1/μ for v to find f1 .

f1=12LT1μ=T14L2μ (I)

Formula to calculate the frequency of wave in second string from equation (I) is,

f2=T24L2μ

• f2 is the frequency in the second string.
• T2 is the tension in the second string.

Multiply and divide by T1 .

f2=T14L2μ(T2T1)=(T14L2μ)T2T1=f1T2T1

Thus, the expression for the frequency of wave in second string is f2=f1T1T1

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