Oscillation of a 600 Hztuning fork sets up standing waves in a string clamped at both ends. The wave speed for the string is 400 m/s. The standing wave has four loops and an amplitude of 2.0 mm. (a) What is the length of the string? (b) Write an equation for the displacement of the string as a function of position and time.

Frequency of the wave, f = 600 Hz

Wave speed, v = 400 m/s

Number of loops, n = 4

Amplitude of the wave,$y_m=2.0mmor2.0\times {10}^{-3}m$

We know the relation between the number of loops and wavelength and also between the wavelength and frequency. So, using these relations, we can find the length of the string.

Then, we use the relation between frequency and angular frequency, wavelength and wavenumber, and amplitude to write the equation for the displacement of the string as a function of position and time.

Formula:

The wavelength of oscillation of n-loops,$\lambda =2L/n........\left(1\right)$

The wavelength of a wave,$\lambda =v/f.......\left(2\right)$

The angular frequency of a wave,$\omega =2\mathrm{\pi f}.....\left(3\right)$

The wavenumber of the wave, $k=2\mathrm{\pi }/\mathrm{\lambda }.......\left(4\right)$

Using equation (1), the wavelength relation to length for n = 4 is given as:

$$\begin{gathered} \lambda =\frac{2L}{4} \\ =\frac{L}{2} \end{gathered}$$

Again, from equation (2), we get the wavelength value and hence equating it to the above equation, we get

So,

$$\begin{gathered} \frac{L}{2}=\frac{v}{f} \\ L=\frac{2v}{f} \\ =\frac{2\left(400m/s\right)}{600Hz} \end{gathered}$$

Therefore, the length of the string is 1.3 m .

The equation of displacement of a resultant wave is given as:

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

But, using equations (4) and (2), we get the value of wavenumber as:

$$\begin{gathered} k=\frac{2\mathrm{\pi f}}{v} \\ =\frac{2\mathrm{\pi }\left(600\mathrm{Hz}\right)}{\left(400m/s\right)} \\ =9.4m^{-1} \end{gathered}$$

And, using equation (3), the value of angular frequency is given as:

$$\begin{gathered} \omega =2\mathrm{\pi }\left(600\mathrm{Hz}\right) \\ =3800\mathrm{rad}/\mathrm{s} \end{gathered}$$

And, the amplitude of the wave is given as;

$y_m=2.00mm$

Therefore, substituting the above values in the equation of wave, we get

$y\left(x,t\right)=\left(2.00mm\right)\sin \left[\left(9.4m^{-1}\right)x\right]\cos \left[\left(3800s^{-1}\right)t\right]$.

Therefore, an equation for the displacement of the string as a function of position and time is$y\left(x,t\right)=\left(2.00mm\right)\sin \left[\left(9.4m^{-1}\right)x\right]\cos \left[\left(3800s^{-1}\right)t\right]$