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14.Waves and Sound
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If the length of a stretched string is shortened by $40\%$ and the tension is increased by $44\%$, then the ratio of the final and initial fundamental frequencies is
A
$2:1$
B
$3:2$
C
$3:4$
D
$1:3$
Solution
(a) Fundamental frequency in case of string is
$n = \frac{1}{{2l}}\sqrt {\frac{T}{m}} \Rightarrow n \propto \frac{{\sqrt T }}{l} \Rightarrow \frac{{n'}}{n} = \sqrt {\frac{{T'}}{T}} \times \frac{l}{{l'}}$
putting $T' = T + 0.44T = \frac{{144}}{{100}}T$ and $l' = l – 0.4l = \frac{3}{5}l$
We get $\frac{{n'}}{n} = \frac{2}{1}$.
Standard 11
Physics
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