Question
A string is stretched between fixed points separated by $$75.0\,cm.$$ It is observed to have resonant frequencies of $$420\,Hz$$ and $$315\,Hz.$$ There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is
A.
$$155\,Hz$$
B.
$$205\,Hz$$
C.
$$10.5\,Hz$$
D.
$$105\,Hz$$
Answer :
$$105\,Hz$$
Solution :
Given $$l = 75\,cm,{f_1} = 420\,Hz\,{\text{and}}\,{f_2} = 315\,Hz.$$
As, two consecutive resonant frequencies for a string fixed at both ends will be
$$\eqalign{
& {f_1} = \frac{{nv}}{{2l}}\,\,{\text{and}}\,\,{f_2} = \frac{{\left( {n + 1} \right)v}}{{2l}} \cr
& \Rightarrow {f_2} - {f_1} = 420 - 315 \cr
& \Rightarrow \frac{{\left( {n + 1} \right)v}}{{2l}} - \frac{{nv}}{{2l}} = 105\,Hz \cr
& \Rightarrow \frac{v}{{2l}} = 105\,Hz \cr} $$
Thus, lowest resonant frequency of a string is $$105\,Hz.$$