14.Waves and Sound
medium

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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