Question

A generator at one end of a very long string creates a wave given by

${\mathit{y}}_{\mathbf{1}}\mathbf{=}\left(6.0cm\right)\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{\pi }}{\mathbf{2}}\left[\left(2.00m^{-1}\right)x+\left(8.00s^{-1}\right)t\right]$

and a generator at the other end creates the wave

${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{)}\mathbf{cos}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{[}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{-}\mathbf{1}}\mathbf{)}\mathbf{x}\mathbf{-}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{)}\mathbf{t}\mathbf{]}$

Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave.

For $\mathit{x}\mathbf{\ge }\mathbf{0}$, what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of x? For$\mathit{x}\mathbf{\ge }\mathbf{0}$, what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of$\mathit{x}$?

Short answer

1. The frequency of each wave will be 2.00 Hz .
2. The wavelength of each wave will be 2.00 m .
3. The speed of each wave will be 4.00 m/s .
4. The location of the node having the smallest value of x will be 0.500 m .
5. The location of the node having the second smallest value of x will be 1.50 m .
6. The location of the node having the third smallest value of x will be 2,50 m.
7. Location of anti-node having smallest value of x will be 0 m .
8. Location of anti-node having second smallest value of x will be 1.00 m .
9. Location of anti-node having third smallest value of x will be 2.00 m .

Step-by-step solution

The given data

The two given wave equations are:

$$\begin{gathered} y_1=(6.0\mathrm{cm})cos\frac{\mathrm{\pi }}{2}[(2.00{\mathrm{m}}^{-1})\mathrm{x}+(8.00{\mathrm{s}}^{-1})\mathrm{t}] \\ {y}_2=(6.0\mathrm{cm})\cos \frac{\mathrm{\pi }}{2}[(2.00{\mathrm{m}}^{-1})\mathrm{x}-(8.00{\mathrm{s}}^{-1})\mathrm{t}] \end{gathered}$$

Understanding the concept of resonant frequency

We are given two different equations of the wave. As there are two generators at both ends, using this, we can find the resultant of two waves. We compare this result to the general equation of the wave. Using this, we can find the required quantities.

Formula:

The general expression of the standing wave,

$y=y_m\cos \left(kx+\omega t\right)$ (i)

The angular frequency of the wave,

$\omega =2\pi f$ (ii)

The wavelength of the wave,

$\lambda =\frac{2\pi }{k}$ (iii)

The velocity of the wave,

$v=f\lambda$ (iv)

Here k is the wave number f, is the frequency of the wave, and$y_m$ is the amplitude of the wave.

a) Calculation of frequency of each wave

We have,

Wave equation generated by the generator at one end,

$y_1=(6.0\mathrm{cm})cos\frac{\mathrm{\pi }}{2}\left((2.00{\mathrm{m}}^{-1})\mathrm{x}+(8.00{\mathrm{s}}^{-1})\mathrm{t}\right)$ (v)

Wave equation generated by the generator at the other end,

$y_2=(6.0\mathrm{cm})cos\frac{\mathrm{\pi }}{2}\left((2.00{\mathrm{m}}^{-1})\mathrm{x}-(8.00{\mathrm{s}}^{-1})\mathrm{t}\right)$ (vi)

By comparing equations (v) and (vi) with equation (i), we can say that the angular frequency of the wave is given as:

role="math" $$\begin{gathered} \omega =\frac{\pi }{2}\left(8.00\right) \\ =4\pi \mathrm{rad}/\mathrm{s} \end{gathered}$$

The frequency of each wave using equation (ii) is given as:

$$\begin{gathered} f=\frac{\omega }{2\pi } \\ =\frac{4\pi }{2\pi } \\ =2.00\mathrm{Hz} \end{gathered}$$

Hence, the frequency of each wave is. 2.00 Hz.

b) Calculation of wavelength of each wave

By comparing equations (v) and (vi) with equation(i), we can say that the wavenumber is given as:

$$\begin{gathered} k=\frac{\pi }{2}\left(2.00\right) \\ =\pi {\mathrm{m}}^{-1} \end{gathered}$$

Using equation (iii), we get the wavelength of each wave as;

$\begin{array}{l}\lambda =\frac{2\mathrm{\pi }}{\mathrm{\pi }}\\ =2.00\mathrm{m}\end{array}$

Hence, the value of the wavelength of each wave is 2.00 m.

c) Calculation of speed of the wave

Using equation (iv), we get the speed of the waves as:

$$\begin{gathered} v=2\times 2 \\ =4.00\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of wave speed of each wave is 4.00 m/s

d) Calculation of location of the node with the smallest value

We know that,

$$\begin{gathered} \cos \alpha +\cos \beta =\sin \left(\alpha +\frac{\pi }{2}\right)+\sin \left(\beta +\frac{\pi }{2}\right) \\ =2\times \sin \left(\frac{\alpha +\beta +\pi }{2}\right)\times \cos \left(\frac{\alpha +\beta }{2}\right) \\ \cos \alpha +\cos \beta =2\times \cos \left(\frac{\alpha +\beta }{2}\right)\times \cos \left(\frac{\alpha +\beta }{2}\right) \end{gathered}$$

Substituting the values $\alpha =kx+\omega t$and $\beta =kx-\omega t$in the above expression, and we get,

$$\begin{gathered} \cos \left(kx+\omega t\right)+\cos \left(kx-\omega t\right)=2\times \cos \left(\frac{\left(kx+\omega t\right)+\left(kx-\omega t\right)}{2}\right)\times \cos \left(\frac{\left(kx+\omega t\right)-\left(kx-\omega t\right)}{2}\right) \\ {y}_m\cos \left(kx+\omega t\right)+y_m\cos \left(kx-\omega t\right)=2\times y_m\cos \left(\frac{\left(2kx\right)}{2}\right)\times \cos \left(\frac{\left(-2\omega t\right)}{2}\right) \\ y=2y_m\cos \left(kx\right)\cos \left(\omega t\right) \end{gathered}$$

We know that nodes occur where,

$\cos \left(kx\right)=0$

Which means,

$kx=n\pi +\frac{\pi }{2}$ (vii)

Where, n = 0,1,2,.....

The smallest value of x will be when n = 0, and that can be calculated as,

$$\begin{gathered} \pi x=0\pi +\frac{\pi }{2} \\ \pi x=\frac{\pi }{2} \\ x=\frac{1}{2} \\ =0.500\mathrm{m} \end{gathered}$$

Hence, the node is located at 0.500 m

e) Calculation of location of the node with the second smallest value

The second smallest value of x will be when n = 1 , and that can be calculated from equation (vii) as,

$$\begin{gathered} \pi x=1\pi +\frac{\pi }{2} \\ \pi x=\pi +\frac{\pi }{2} \\ \pi x=\frac{3\pi }{2} \\ x=\frac{3}{2} \\ =1.50\mathrm{m} \end{gathered}$$

Hence, the node is located at 1.50 m .

f) Calculation of location of node at the third smallest value

The second smallest value of x will be when n = 2 , and that can be calculated from equation (vii) as,

$$\begin{gathered} \pi x=2\pi +\frac{\pi }{2} \\ \pi x=2\pi +\frac{\pi }{2} \\ \pi x=\frac{5\pi }{2} \\ x=\frac{5}{2} \\ =2.50\mathrm{m} \end{gathered}$$

Hence, the node is located at 2.50 m

g) Calculation of anti-node with the smallest value

Anti-nodes occur when,

$\cos \left(kx\right)=\pm 1$

Which means,

$kx=n\pi$ (vii)

Where, n = 0,1,2,.......

The smallest value of x will be when n = 0 , and that can be calculated from equation (viii) as,

$$\begin{gathered} \pi x=0\pi \\ x=0.0m \end{gathered}$$

Hence, the anti-node is located at 0 m .

h) Calculation of anti-node with the second smallest value

For the second smallest value of x will be when n = 1 , and that can be calculated from equation (viii) as,

$$\begin{gathered} \pi x=1\pi \\ x=1.00m \end{gathered}$$

Hence, the anti-node is located at 1.00 m .

i) Calculation of anti-node with the third smallest value

For the second smallest value of x will be when n = 2 , and that can be calculated from equation (viii) as,

$$\begin{gathered} \pi x=2\pi \\ x=2.00\mathrm{m} \end{gathered}$$

Hence, the anti-node is located at 2.00 m .