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Wize University Physics Textbook (Master) > Waves: Mechanical

Standing Waves in Strings with Both Ends Fixed

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Standing Waves in Strings with Both Ends Fixed


In order to see a standing wave on a string with both ends fixed, the wavelength (and therefore frequency) must have a certain relation with the length of the string, leading to specific frequencies for a given string.


Before finding these specific frequencies is good to remember that:

  • There are always nodes at the fixed ends of the string.
  • The distance between any two successive nodes or successive antinodes is λ2\dfrac{\lambda}{2}.
  • The distance between any two successive node and antinodes is λ4\dfrac{\lambda}{4}.












Let's consider a string with length LL





Hence, the length of the string fixed at both ends could be any integer multiple of λ2\dfrac{\lambda}{2}

L=n⋅λ2n=1,2,3,…or λn=2Lnn=1,2,3,…\begin{array}{l}L=n\cdot\dfrac{\lambda}{2}\quad n=1,2,3,\ldots\\ \\\text{or }\lambda_n=\dfrac{2L}{n}\quad n=1,2,3,\ldots\end{array}

PAGE BREAK


Using v=λfv=\lambda f, the corresponding frequencies are:


fn=vλn=nv2Ln=1,2,3,…\boxed{f_n=\dfrac{v}{\lambda_n}=n\dfrac{v}{2L}\quad n=1,2,3,\ldots}




The smallest frequency creating a standing wave along a sting is known as the fundamental frequency and it is shown by f1f_1 :


f1=v2L\boxed{f_1=\dfrac{v}{2L}}




fnf_n's are called resonant series or harmonic series and fn(n>1)f_n(n>1)are also called overtones. These resonant frequencies are related to the fundamental frequency as:



fn=nf1   n=1,2,3,...\boxed{f_n=nf_1 \ \ \ n=1,2,3,...}


Exam Tip
For strings fixed at both ends, the difference between any two successive resonant frequencies is equal to the fundamental frequency.





Example: Standing Wave Along a String Fixed at Both Ends


A series of resonance patterns is measured for a 8080cm string fixed at both ends. The successive resonant frequencies are reported to be 16021602, 18691869 and 21362136 Hz.

a) What is the fundamental frequency of this string?
b) If the mass density of the string is 55 g/m,, what is the wave speed and the tension along the string?
c) Draw the resonance pattern corresponding to 13351335 Hz.

Part a)

For strings with both ends fixed, the difference between any two successive frequencies is the fundamental frequency:


fn+1−fn=(n+1)f1−nf1f_{n+1}-f_n=\left(n+1\right)f_1-nf_1

=nf1+f1−nf1=nf_1+f_1-nf_1

=f1=f_1

Hence:

f1=1869−1602=267f_1=1869-1602=267 (Hz)


Part b)

The fundamental frequency of a string fixed at both ends is related to the wave speed as:

f1=v2Lf_1=\dfrac{v}{2L}

Solve for the speed:

v=2Lf1 v=2Lf_1

=2(0.8)(267)=427.2=2(0.8)(267)=427.2 (m/s)


We also know that the wave speed in strings depends on their tension and mass density:

v=FTμ   →   v2=FTμ   →   FT=μv2v=\sqrt{\dfrac{F_T}{\mu}} \ \ \ \to \ \ \ v^2=\dfrac{F_T}{\mu} \ \ \ \to \ \ \ F_T=\mu v^2


Therefore the force of tension is:

FT=(0.005)(427.2)2=912.5F_T=(0.005)(427.2)^2=912.5 (N)


Part c)

First, let's find the resonance number (nn) corresponding to the given frequency fn=1335f_n=1335:

fn=nf1   →   n=fnf1=1335267=5f_n=nf_1 \ \ \ \to \ \ \ n=\dfrac{f_n}{f_1}=\dfrac{1335}{267}=5

This means the pattern has 5 half wavelengths:



Practice: String Tension and Fundamental Frequency


Standing waves are observed in a string with length of 2020 cm. If the tension along the string is increased by a factor of 99, what is the new fundamental frequency of the wave compared to its original value?