Question

A uniform string resonates with a tuning fork, at a maximum tension of $32N$. 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 $l$.
$f=\sqrt{\frac{T}{\mu }}\times \frac{1}{2l}\dots$ (i)
$f=\sqrt{\frac{T}{\mu }}\times \frac{4}{2l}\dots$(ii)
or $f=\sqrt{\frac{T}{\mu }}\times \frac{4}{61}\dots$ (iii)
Equating (i) $\&$ (ii) and (i) $\&$ (iii)
$\sqrt{\frac{T}{T_1}}=4\&\sqrt{\frac{T}{T_2}}=\frac{4}{3}$
Put $T=32N$
$\frac{32}{16}=T_1$
$\frac{9}{16}\times 32=T_2$
$T_1=2N$
$T_2=18N$
of the options on $T_1$ is right.