Physics - Wave on String Question with Solution | TestHub

The frequency of the $p^{\text{th}}$ harmonic of a stretched string fixed at both ends is given by $f_p=p{f}_1$, where $f_1$ is the fundamental frequency.

Given that the wire resonates in the $p^{\text{th}}$ harmonic at $320\text{ Hz}$:
$p{f}_1=320$

The next higher resonant frequency corresponds to the $(p+1{)}^{\text{th}}$ harmonic, which is $400\text{ Hz}$:
$(p+1)f_1=400$

Subtracting the first equation from the second gives the fundamental frequency:
$(p+1)f_1-p{f}_1=400-320$
$f_1=80\text{ Hz}$

Substituting the value of $f_1$ back into the first equation:
$p\times 80=320$
$p=\frac{320}{80}=4$

Answer: 4