$A$ tuning fork gives $5 \text{ beats per second}$ with a $33 \text{ cm}$ length of sonometer wire. If the length of the wire is shortened by $1 \text{ cm}$, the number of beats is still the same. The frequency of the fork is (in $\text{ Hz}$)

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:

$A$ sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \ kg$ is suspended from the wire. When this mass is replaced by a mass $M$,the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is ... $kg$.

If you set up the $7^{th}$ overtone on a string fixed at both ends,how many nodes and antinodes are set up in it?

When tension $T$ is applied to a sonometer wire of length $l$, it vibrates with the fundamental frequency $n$. Keeping the setup same, when the tension is increased by $8 \,N$, the fundamental frequency becomes three times the earlier. The initial tension applied to the wire was: (in $\,N$)

The fundamental frequency of a sonometer wire is $50 \text{ Hz}$ for a given length and tension. If the length is increased by $25\%$ while keeping the tension constant, what is the percentage change in the frequency of the second harmonic?