Question

Physics - Standing wave modes on a string - assignment

Answer 1

a)

When a standing wave propagates through a string, the length of the string is a multiple of half the wavelength:

$L=n\cdot\frac{\lambda_n}{2}$

The factor n corresponds to the number of the harmonic. Then, the first harmonic is given by the condition n=1:

$\begin{gathered} L=\frac{\lambda_1}{2} \\ \Rightarrow\lambda_1=2L \end{gathered}$

As we can see, the wavelength of the first harmonic is two times the length of the string.

Then, the wavelength of the first harmonic can be found by replacing the length L=63cm:

$\lambda_1=2\times63cm=126cm$

b)

The product of the wavelength and the frequency is the speed of the wave:

$v=\lambda f$

Replace λ=126cm=1.26m and f=330Hz to find the speed of the wave on the E-string:

$v=(1.26m)(330Hz)=415.8\frac{m}{s}\approx416\frac{m}{s}$

c)

The frequency of the n-th harmonic is given by:

$f_n=\frac{v}{\lambda_n}$

On the other hand:

$\lambda_n=\frac{2L}{n}$

Then:

$f_n=\frac{v}{2L}\times n$

Notice that v/2L is the frequency of the first harmonic (fundamental frequency). Then:

$f_n=f_1\times n$

Replace the fundamental frequency of 330Hz and n=2,3,4 to find the second, third and fourth harmonic frequencies:

$\begin{gathered} f_2=330Hz\times2=660Hz \\ f_3=330Hz\times3=990Hz \\ f_4=330Hz\times4=1320Hz \end{gathered}$

d)

Replace n=3 into the expression for the wavelength of the n-th harmonic to find the wavelength of the third harmonic:

$\lambda_n=\frac{2L}{n}=\frac{2\times63cm}{3}=\frac{126cm}{3}=42cm$

Therefore, the answers are:

a) 126cm

b) 416m/s

c) 660Hz, 990Hz, 1320Hz

d) 42cm