heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end of the wire, a pulse travels down the wire and returns 7.84 ms later, having reflected off the lower end. The speed of sound in the room is 344 m/s, and the stone is heavy enough to prevent the lower end of the wire from moving. If the wire is vibrating in its second overtone, what is the wavelength of the sound it will produce?

The equation that relates frequency and wavelength of the sound produced is [tex]\lambda=\frac{c}{f}[/tex], where c is the speed of sound in the room in this case. First we must know the frequency at which the string is vibrating.

The equation that relates frequency on a string with its longitude L and pulse speed v while vibrating on the armonic n is:

[tex]f_n=\frac{nv}{2L}[/tex]

The velocity of the pulse v is the length of the string L divided by the time it takes the pulse to travel the string. The time given t is the time to go forth and back, so the time to travel the string once is half that. We have then:

[tex]v=\frac{L}{t/2}=\frac{2L}{t}[/tex]

We must put all together now:

[tex]f_n=\frac{nv}{2L}=\frac{n\frac{2L}{t}}{2L}=\frac{n}{t}[/tex]

Since the frequency of the sound produced f is the frequency of the vibrating string [tex]f_n[/tex], and taking into account that the 2nd overtone is the 3rd armonic, we finally have:

[tex]\lambda=\frac{c}{f}=\frac{c}{\frac{n}{t}}=\frac{ct}{n}=\frac{(344m/s)(7.84\times10^{-3}s)}{3}=0.9m[/tex]