Q.8. If the length of stretched string is reduced by 40% and tension is increased by 44% then the ratio of final to initial frequencies of stretched string is
A. 2:1
B. 3:2
C. 3:4
D. 1:3
Answer :- A. 2:1
Explanation :-
Let the initial length and tension be l and T respectively.
The new length l_{new}=l-\frac{40}{100}l=\frac{3}{5}l
After increase in tension,
the new tension T_{new}=T+\frac{44}{100}T=\frac{144T}{100}
Fundamental frequency of a vibrating string is given by
\mathrm{n} =\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}}} \therefore \quad \mathrm{n}_1 =\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}}} \mathrm{n}_2 =\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\text {new }}}{\mathrm{m}}} \therefore \quad \frac{\mathrm{n}_1}{\mathrm{n}_2} =\frac{l_{\text {new }}}{l} \times \frac{\sqrt{\mathrm{T}}}{\sqrt{\mathrm{~T}_{\text {new }}}} =\frac{\frac{3}{5} l}{l} \times \sqrt{\frac{100 \mathrm{~T}}{144 \mathrm{~T}}} =\frac{3}{5} \times \frac{10}{12}=\frac{1}{2} \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1} =\frac{2}{1}