To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
![\lambda = (v)/(f)](https://img.qammunity.org/2020/formulas/physics/high-school/ukyaqbtm4hhypaise2zv0rzg78a2xj4rww.png)
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
![\lambda = (192)/(320)](https://img.qammunity.org/2020/formulas/physics/college/1m0wnsutl1uqw2tair5en8yi7u9gatyhy1.png)
![\lambda = 0.6m](https://img.qammunity.org/2020/formulas/physics/college/znfzg9aw073enl8mi8kyqjjmbrvfnkfw71.png)
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
![N = (L)/(\lambda)](https://img.qammunity.org/2020/formulas/physics/high-school/h1l0nxz330njz851sb341kklq2xh5n6uml.png)
![N = (3.6)/(0.6)](https://img.qammunity.org/2020/formulas/physics/college/brcpycuc3ddtu581i6sau8gk9e08drw5o2.png)
![N = 6](https://img.qammunity.org/2020/formulas/physics/college/ttgoeyza3hwjk40vwcosw0359x2i5oj5a3.png)
Therefore the number of wavelengths of the wave fit on the string is 6.