Physics - Wave on String Question with Solution | TestHub

Let the fundamental frequency of the stretched wire be $n_1$ and the fundamental frequency of the closed pipe be $n_c$.

$n_1=\frac{1}{2L}\sqrt{\frac{T}{m}}$

$n_c=\frac{v}{4L}$

Given that the wire is in resonance with the fundamental frequency of the closed pipe:

$n_1=n_c\Rightarrow \frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{v}{4L}$

When the tension is increased by $8$ N, the new frequency of the wire is $n_2$:

$n_2=\frac{1}{2L}\sqrt{\frac{T+8}{m}}$

The first overtone of a closed pipe is $3n_c$. The wire is now in resonance with this first overtone:

$n_2=3n_c\Rightarrow \frac{1}{2L}\sqrt{\frac{T+8}{m}}=3\left(\frac{v}{4L}\right)$

Dividing the equation for $n_2$ by the equation for $n_1$:

$\sqrt{\frac{T+8}{T}}=3$

Squaring both sides:

$\frac{T+8}{T}=9$

$T+8=9T$

$8T=8\Rightarrow T=1$ N

Answer: 1 N