Question 393: Waves

Question

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

Accepted Answer

Let the initial length and tension be $$l$$ and $$T$$ respectively.
After shortening,
The new length $$l_{	ext {new }}=l-\frac{40}{100} l=\frac{3}{5} l$$
After increase in tension,
the new tension $$\mathrm{T}_{	ext {new }}=\mathrm{T}+\frac{44}{100} \mathrm{~T}=\frac{144 \mathrm{~T}}{100}$$
Fundamental frequency of a vibrating string is given by
$$\begin{aligned}
\mathrm{n} & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \
	herefore \quad \mathrm{n}_1 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \
\mathrm{n}_2 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{	ext {new }}}{\mathrm{m}}} \
	herefore \quad \frac{\mathrm{n}_1}{\mathrm{n}_2} & =\frac{l_{	ext {new }}}{l} 	imes \frac{\sqrt{\mathrm{T}}}{\sqrt{\mathrm{T}_{	ext {new }}}} \
& =\frac{\frac{3}{5} l}{l} 	imes \sqrt{\frac{100 \mathrm{~T}}{144 \mathrm{~T}}} \
& =\frac{3}{5} 	imes \frac{10}{12}=\frac{1}{2} \
	herefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1} & =\frac{2}{1}
\end{aligned}$$

Answer

$$2: 1$$

Answer

$$3: 2$$

Answer

$$3: 4$$

Answer

$$1: 3$$

Waves

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

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