JEE PYQ: Waves And Sound Question 23
Question 23 - 2020 (09 Jan Shift 2)
A wire of length L and mass per unit length $6.0 \times 10^{-3}$ kgm$^{-1}$ is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is:
(1) 2.1 m
(2) 1.1 m
(3) 8.1 m
(4) 5.1 m
Show Answer
Answer: (1)
Solution
Fundamental frequency, $f = 70$ Hz.
The fundamental frequency of wire vibrating under tension $T$ is given by
$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
Here, $\mu =$ mass per unit length of the wire, $L =$ length of wire
$70 = \frac{1}{2L}\sqrt{\frac{540}{6 \times 10^{-3}}}$
$\Rightarrow L = 2.14$ m
BETA
