Question 340: Waves

Question

A violin emits sound waves of frequency ' $n_1$ ' under tension T. When tension is increased by $44 \%$, keeping the length and mass per unit length constant, frequency of sound waves becomes ' $\mathrm{n}_2$ '. The ratio of frequency ' $\mathrm{n}_2$ ' to frequency ' $n_1$ ' is

Accepted Answer

The frequency of a vibrating string is given by the formula:
$$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$
where:
$ n $ is the frequency,
$ L $ is the length of the string,
$ T $ is the tension in the string,
$ \mu $ is the mass per unit length of the string.
Given that the initial frequency is $ n_1 $ with tension $ T $, when the tension is increased by $ 44\% $, the new tension becomes:
$$ T' = T + 0.44T = 1.44T $$
The new frequency $ n_2 $ when the tension is $ T' $ is:
$$ n_2 = \frac{1}{2L} \sqrt{\frac{1.44T}{\mu}} $$
We can simplify this to:
$$ n_2 = \frac{1}{2L} \cdot \sqrt{\frac{1.44T}{\mu}} = \frac{1}{2L} \cdot \sqrt{1.44} \cdot \sqrt{\frac{T}{\mu}} $$
Recognizing that $ \sqrt{1.44} = 1.2 $, we have:
$$ n_2 = 1.2 \cdot \frac{1}{2L} \cdot \sqrt{\frac{T}{\mu}} $$
Since we know $ n_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $, we can substitute:
$$ n_2 = 1.2 \cdot n_1 $$
The ratio of $ n_2 $ to $ n_1 $ is:
$$ \frac{n_2}{n_1} = 1.2 $$
Expressing $ 1.2 $ as a ratio, we have:
$$ \frac{n_2}{n_1} = \frac{6}{5} $$
Therefore, the correct answer is:
Option C: $6: 5$

Answer

$5: 6$

Answer

$6: 7$

Answer

$6: 5$

Answer

$7: 6$

Waves

A violin emits sound waves of frequency ' n_1 ' under tension T. When tension is increased by 44 \%, keeping the length and mass per unit length constant, frequency of sound waves becomes ' \mathrm{n}_2 '. The ratio of frequency ' \mathrm{n}_2 ' to frequency ' n_1 ' is

Choose the correct answer: