$A$ string of length $1\, m$ and mass $5\, g$ is fixed at both ends. The tension in the string is $8.0\, N$. The string is set into vibration using an external vibrator of frequency $100\, Hz$. The separation between successive nodes on the string is close to .... $cm$

$A$ uniform string resonates with a tuning fork at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of the length from one end,then to resonate with the same frequency,the maximum value of tension for the string will be ........... $N$.

$A$ sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same,the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)

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 $\,N$)

$A$ segment of wire vibrates with a fundamental frequency of $450 \,Hz$ under a tension of $9 \,kg-wt$. The tension at which the fundamental frequency of the same wire becomes $900 \,Hz$ is: