Q. The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by $8\ N,$ it is in resonance with the first overtone of the closed tube. The initial tension in the wire is
A) $1\ N$
B) $4\ N$
C) $8\ N$
D) $16\ N$
Explanation
(a) According to problem
$\frac{1}{2L}\sqrt{\frac{T}{m}} = \frac{v}{4L}$ …(i)
And $\frac{1}{2L}\sqrt{\frac{T + 8}{m}} = \frac{3v}{4L}$ …(ii)
Dividing equation (i) and (ii),
$\sqrt{\frac{T}{T + 8}} = \frac{1}{3} \Rightarrow T = 1N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.