$A$ stretched string of $1 \ m$ length and mass $5 \times 10^{-4} \ kg$ is having a tension of $20 \ N$. If it is plucked at $25 \ cm$ from one end,then it will vibrate with a frequency of ... $Hz$.

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?

$A$ string vibrates in its fundamental mode when a tension $T_1$ is applied to it. If the length of the string is decreased by $25 \%$ and the tension applied is changed to $T_2$, the fundamental frequency of the string increases by $100 \%$, then $\frac{T_2}{T_1} =$ (Linear density of the string is constant)

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

The correct graph between the frequency $n$ and the square root of density $(\rho)$ of a wire,keeping its length,radius,and tension constant,is:

$A$ 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 .... $cm$