Q38E - University Physics with Modern Physics

Question

 \(1.50\;{\rm{m}}\)-long rope is stretched between two supports with a tension that makes the speed of transverse waves \(62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\). What are the wavelength and frequency of 

(a) the fundamental;

 (b) the second overtone;

 (c) the fourth harmonic?

Step-by-Step Solution

Answer

(a) The wavelength and frequency of the fundamental are, \(3.0\;{\rm{m}}\) and \(20.67\;{\rm{Hz}}\).

Step 1: Identification of the given data

The given data can be listed below as,

• The length of the rope is, \(L = 1.50\;{\rm{m}}\).
• The value of \(n\) is, 1.
• The speed of transverse wave is, \(v = 62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

Step 2: Concept

The relationship between frequency, wavelength, and speed as well as the condition for harmonics are the major concepts employed to address this problem. Use the fundamental tone condition to determine wavelength first, and then the relationship between frequency, wavelength, and speed to determine frequency.

 

The following expression is used to determine the wavelength of various tones:

\(\lambda=\frac{{2L}}{n}\)                   …(1)

 

Here \(\lambda \) is the wavelength, \(L\) is the length, and \(n\) is the number of tone.

Step 3: Determination of the wavelength and frequency of the fundamental

The wavelength and frequency of the fundament is expressed as,

 

The wavelength of various tones is calculated using the expression as follows:

\(\lambda  = \frac{{2L}}{n}\) 

Substitute \(1.50\;{\rm{m}}\) for \(L\) and 1 for \(n\) in the above equation (1)

\(\begin{array}{c}\lambda  = \frac{{2 \times 1.50\;{\rm{m}}}}{1}\\\lambda  = 3.0\;{\rm{m}}\end{array}\) 

Hence the wavelength is, \(3.0\;{\rm{m}}\).

The relation between frequency, wavelength, and speed is expressed as,

\(f=\frac{v}{\lambda}\)                        …(2)

 

Here \(f\) is the frequency, \(v\) is the speed of transverse wave and \(\lambda \) is wavelength. 

Substitute \(62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) for \(v\) and \(3.0\;{\rm{m}}\) for \(\lambda \) in the above equation (2)

\(\begin{array}{c}f = \frac{{62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}}}{{3.0\;{\rm{m}}}}\\f = 20.67\;{\rm{Hz}}\end{array}\)

Hence the frequency is, \(20.67\;{\rm{Hz}}\). 

Therefore, the wavelength and frequency of the fundamental are, \(3.0\;{\rm{m}}\) and \(20.67\;{\rm{Hz}}\).