Chapter 12, Problem 47P

(a)

To determine

The frequency of second piano.

Answer to Problem 47P

  $f_s=132.5\text{ Hz}$ or $130.5\text{ Hz}.$

Explanation of Solution

Given:

Frequency of first piano strings $\left(f_f\right)=132\text{Hz}.$

Tuner hears 3 beats at every $2\text{ s}\text{.}$

Formula used:

  $f_s=f_f+f_B$ and $f_s=f_f-f_B$

Where,

  $f_s$ = frequency of second piano, Hz

  $f_f$ = frequency of first piano, Hz

  $f_B$ = Beat frequency, Hz.

Calculation:

Vibration of two piano strings are $132\text{ Hz}.$

Frequency of beats hear by the tuner is 3 beats per 2 seconds that is $\frac{3\text{ beats}}{2\text{ s}}=1.5\text{beats/s}$ .

Now, determine the frequency of second piano string,

  $f_s=f_f+f_B$

  $f_s=132+1.5=133.5\text{ Hz}$

Also,

  $\begin{array}{l}{f}_s=f_f-f_B\\ f_s=130.5\text{ Hz}\text{.}\end{array}$

The frequency of the second beat is $132.5\text{ Hz}\text{.}$ or $130.5\text{ Hz}.$

Conclusion:

The frequency of the second piano string is $132.5\text{Hz}$ or $130.5\text{ Hz}.$

(b)

To determine

The percentage of tension that is increased or decreased to bring them in tune.

Answer to Problem 47P

Percentage of tension that is decreased by $4.5\%$ and increased by $4\%.$

Explanation of Solution

Given:

Frequency of first piano strings $\left(f_f\right)=132\text{Hz}\text{.}$

Tuner hears 3 beats at every $2\text{ s}.$

Formula used:

  $\sqrt{\frac{T_{is}}{T_{os}}}=\frac{f_{is}}{f_{os}}$

Where,

  $T_{is}$ = Tension in first piano.

  $T_{os}$ = Tension in second piano.

  $f_{is}$ = frequency in first piano, Hz

  $f_{os}$ = frequency in second piano, Hz

Calculation:

Since the strings are of same length, the wavelength is same, so the ratio of tensions for second piano string is given as,

  $\sqrt{\frac{T_{is}}{T_{os}}}=\frac{f_{is}}{f_{os}}$

  $\sqrt{\frac{T_{is}}{T_{os}}}=0.977$

  $\begin{array}{l}$ \frac{T_{is}}{T_{os}}={(0.977)}^2$\frac{T_{is}}{T_{os}}=0.954\end{array}$

  $\frac{T_{os}-T_{is}}{T_{os}}=0.045$$$

  $\frac{T_{os}-T_{is}}{T_{os}}=4.5\%$ (decreased)

It is decreased by 4.5%.

  $\frac{T_{os}-T_{is}}{T_{os}}=4.5\%(\text{decreased)}$

Similarly,

  $\sqrt{\frac{T_{is}}{T_{os}}}=\frac{f_{is}}{f_{os}}$

  $\sqrt{\frac{T_{is}}{T_{os}}}=1.02$

  $\begin{array}{l}$ 1-\frac{T_{is}}{T_{os}}={(1-1.02)}^2$1-\frac{T_{is}}{T_{os}}=0.40\end{array}$

  $\frac{T_{os}-T_{is}}{T_{os}}=0.04$

  $\frac{T_{os}-T_{is}}{T_{os}}=4\%$ (increased)

It is increased by 4%.

Conclusion:

The percentage of the tension is increased by 4% or decreased by 4.5%.