Q47E - University Physics with Modern Physics

Question

Guitar String. One of the \(63.5\;{\rm{cm}}\)-long strings of an ordinary guitar is tuned to produce the note \({B_3}\) (frequency \(245\;{\rm{Hz}}\)) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string.

 (b) If the tension in this string is increased by \(1.0\% \), what will be the new fundamental frequency of the string?

 (c) If the speed of sound in the surrounding air is \(344\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\), find the frequency and wavelength of the sound wave produced in the air by the vibration of the \({B_3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

Step-by-Step Solution

Answer

(a) The speed of transverse waves is, \(311.15\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

Step 1: Identification of the given data

The given data can be listed below as,

• The length of the string is, \(L = 63.5\;{\rm{cm}} = 63.5 \times {10^{ - 2}}\;{\rm{m}}\).
• The fundamental frequency is, \({f_1} = 245\;{\rm{Hz}}\).

Step 2: Significance of the speed of transverse waves

A wave's velocity is its rate of movement. By using the formula wave speed is equal to the product of frequency and wavelength, wave speed is correlated with wavelength, frequency, and period. Visible light, an electromagnetic wave, travels at a speed that is most frequently employed.

Step 3: Determination of the speed of transverse waves

The relation between fundamental frequency, velocity and length of string is expressed as,

\({f_1} = \frac{v}{{2L}}\)                                                                                                                        …(1)

Here \({f_1}\) is the fundamental frequency, \(v\) is the velocity and \(L\) is the length of string.

 

Substitute \(245\;{\rm{Hz}}\) for \({f_1}\) and \(63.5 \times {10^{ - 2}}\;{\rm{m}}\) for \(L\) in the equation (1).

\(\begin{array}{c}245\;{\rm{Hz}} = \frac{v}{{2 \times 63.5 \times {{10}^{ - 2}}\;{\rm{m}}}}\\v = \left( {245\;{\rm{Hz}}} \right) \times \left( {2 \times 63.5 \times {{10}^{ - 2}}\;{\rm{m}}} \right)\\v = 311.15\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\}{\rm{s}}}\end{array}\)

 

Hence the speed of transverse waves is, \(311.15\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).