Calculate the frequency of the second harmonic formed on a string of length $0.5 \ m$ and mass $2 \times 10^{-4} \ kg$ when stretched with a tension of $20 \ N$.

At the poles,a stretched wire of a given length vibrates in unison with a tuning fork. At the equator,for the same setting to produce resonance with the same fork,the vibrating length of the wire

$A$ string is stretched between two rigid supports separated by $75 \,cm$. There are no resonant frequencies between $420 \,Hz$ and $315 \,Hz$. The lowest resonant frequency for the string is (in $\,Hz$)

In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is

$A$ string is stretched between two fixed points separated by $75\,cm$. It is observed to have resonant frequencies of $420\,Hz$ and $315\,Hz$. There are no other resonant frequencies between these two. Then,the lowest resonant frequency for this string is ..... $Hz$.

The length of the string of a musical instrument is $90 \;cm$ and has a fundamental frequency of $120 \;Hz$. Where (in $cm$) should it be pressed to produce a fundamental frequency of $180 \;Hz$?