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An organ is in tune at \({\bf{22}}{\bf{.}}{{\bf{0}}^{\bf{o}}}{\bf{C}}\). By what percent will the frequency be off at \({\bf{1}}{{\bf{1}}^{\bf{o}}}{\bf{C}}\)?

Short Answer

Expert verified

The percentage frequency be off at \(11{\rm{^\circ C}}\) is \( - 1.917\% \).

Step by step solution

01

Relationship between the speed of sound, frequency, and temperature.

The speed of sound is directly related to the change in the temperature, and this change in temperature is directly related to the change in frequency.\(v \propto T \propto f\).

02

Calculation of change in frequency

The expression for the frequency at\(22{\rm{^\circ C}}\)is given as,

\({f_{22{\rm{^\circ C}}}} = \frac{{{v_{22{\rm{^\circ C}}}}}}{\lambda }\)

Here,\({v_{22^\circ C}}\)is the speed of the sound at\(22{\rm{^\circ C}}\)and\(\lambda \)is the resonant wavelength inside the pipe.

The expression for the frequency at\(11{\rm{^\circ C}}\)is given as,

\({f_{{\rm{11^\circ C}}}} = \frac{{{v_{{\rm{11^\circ C}}}}}}{\lambda }\)

Here,\({v_{11^\circ C}}\)is the speed of the sound at\(11{\rm{^\circ C}}\).

The expression for the change in frequency is given as,

\(\Delta f = {f_{11{\rm{^\circ C}}}} - {f_{{\rm{22^\circ C}}}}\)

Substitute the values in the above equation,

\(\begin{array}{c}\Delta f = \frac{{{v_{{\rm{11^\circ C}}}}}}{\lambda } - \frac{{{v_{{\rm{22^\circ C}}}}}}{\lambda }\\ = \frac{{{v_{{\rm{11^\circ C}}}} - {v_{{\rm{22^\circ C}}}}}}{\lambda }\end{array}\)

03

Calculation of the percentage change in the frequency.

The ratio of change in frequency to the original frequency is given as,

\(\begin{array}{c}\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \frac{{\frac{{{v_{{\rm{11^\circ C}}}} - {v_{{\rm{22^\circ C}}}}}}{\lambda }}}{{\frac{{{v_{{\rm{22^\circ C}}}}}}{\lambda }}}\\ = \frac{{{v_{{\rm{11^\circ C}}}}}}{{{v_{{\rm{22^\circ C}}}}}} - 1\end{array}\)…… (i)

The speed of sound at\(22{\rm{^\circ C}}\)can be written as,

\({v_{22{\rm{^\circ C}}}} = 331 + 0.6{T_{22{\rm{^\circ C}}}}\)

The speed of sound at\(11{\rm{^\circ C}}\)can be written as,

\({v_{{\rm{11^\circ C}}}} = 331 + 0.6{T_{{\rm{11^\circ C}}}}\)

Substitute the values in equation (i),

\(\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \frac{{331 + 0.6{T_{{\rm{11^\circ C}}}}}}{{331 + 0.6{T_{{\rm{22^\circ C}}}}}} - 1\)

The percentage difference is,

\(\begin{array}{c}\frac{{\Delta f}}{{{f_{22^\circ {\rm{C}}}}}} = \left( {\frac{{331 + 0.6\left( {11{\rm{^\circ C}}} \right)}}{{331 + 0.6\left( {{\rm{22^\circ C}}} \right)}} - 1} \right) \times 100\% \\ = - 0.01917 \times 100\% \\ = - 1.917\% \end{array}\)

Thus, the percentage frequency be off at \(11{\rm{^\circ C}}\) is \( - 1.917\% \).

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Most popular questions from this chapter

A musical note that is two octaves higher than a second note

(a) has twice the frequency of the second note.

(b) has four times the frequency of the second note.

(c) has twice the amplitude of the second note.

(d) is 3 dB louder than the second note.

(e) None of the above.

When a sound wave passes from air into water, what properties of the wave will change?

(a) Frequency.

(b) Wavelength.

(c) Wave speed.

(d) Both frequency and wavelength.

(e) Both wave speed and wavelength.

Question: A bat emits a series of high-frequency sound pulses as it approaches a moth. The pulses are approximately \({\bf{70}}{\bf{.0}}\,{\bf{ms}}\) apart, and each is about \({\bf{3}}{\bf{.0}}\,{\bf{ms}}\) long. How far away can the moth be detected by the bat so that the echo from one pulse returns before the next pulse is emitted?

Question: (II) Two violin strings are tuned to the same frequency, 294 Hz. The tension in one string is then decreased by 2.5%. What will be the beat frequency heard when the two strings are played together? (Hint: Recall Eq. 11–13.)

Question: In audio and communications systems, the gain, \(\beta \) in decibals is defined for an amplifer as,

\(\beta = {\bf{10log}}\left( {\frac{{{P_{{\bf{out}}}}}}{{{P_{{\bf{in}}}}}}} \right)\)

Where \({P_{{\bf{in}}}}\) is the power input to the system and \({P_{{\bf{out}}}}\) is the power output. (a) A particular amplifer puts out 135 W of power for and input of 1.0 mW. What is its gain in dB? (b) if a signal to noise ratio of 93 dB is specified, what is the noise power if the output signal is 10 W?

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