Question

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 length from one end, then to resonance with same frequency the maximum value of tension for string will be

• A. 2 N
• B. 4 N
• C. 8 N
• D. 16 N

Answer 1

Answer: (A)

Let length be $l$.
$f=\sqrt{\frac{T}{\mu}} \times \frac{1}{2 l} \ldots$ (i)
$ f=\sqrt{\frac{T}{\mu}} \times \frac{4}{2 l} \ldots $(ii)
or $ f=\sqrt{\frac{T}{\mu}} \times \frac{4}{61} \ldots $ (iii)
Equating (i) $\&$ (ii) and (i) $\&$ (iii)
$\sqrt{\frac{T}{T_{1}}}=4 \,\& \, \sqrt{\frac{T}{T_{2}}}=\frac{4}{3}$
Put $T=32 \,N$
$\frac{32}{16}=T_{1}$
$ \frac{9}{16} \times 32=T_{2}$
$T_{1}=2 \,N $
$ T_{2}=18 \,N$
of the options on $T_{1}$ is right.