Question 327: Waves

Question

When the tension in string is increased by $3 \mathrm{~kg} \omega \mathrm{t}$, the frequency of the fundamental mode increases in the ratio $2: 3$. The initial tension in the string is

Accepted Answer

For a string under tension, the frequency of the fundamental mode is given by the formula:
$$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$
where:
$ f $ is the frequency,
$ L $ is the length of the string,
$ T $ is the tension in the string,
$ \mu $ is the linear mass density.
Initially, let the tension be $ T $ and the initial frequency be $ f_1 $. When the tension is increased by $ 3 \, \mathrm{kg} \, \omega \mathrm{t} $, the new tension becomes $ T + 3 \, \mathrm{kg} \, \omega \mathrm{t} $, and the new frequency is $ f_2 $.
The ratio of the frequencies is given as $ \frac{f_1}{f_2} = \frac{2}{3} $.
Given this relationship:
$$ \frac{f_2}{f_1} = \frac{3}{2} $$
Using the formula for frequency and the fact that frequency is proportional to the square root of tension:
$$ \frac{f_2}{f_1} = \sqrt{\frac{T + 3}{T}} $$
Equating the ratios:
$$ \frac{3}{2} = \sqrt{\frac{T + 3}{T}} $$
Squaring both sides:
$$ \left(\frac{3}{2}\right)^2 = \frac{T + 3}{T} $$
$$ \frac{9}{4} = \frac{T + 3}{T} $$
Cross-multiplying gives:
$$ 9T = 4(T + 3) $$
Expanding and simplifying:
$$ 9T = 4T + 12 $$
$$ 5T = 12 $$
Thus, the initial tension $ T $ is:
$$ T = \frac{12}{5} = 2.4 \, \mathrm{kg} \, \omega \mathrm{t} $$
Therefore, the initial tension in the string is $ 2.4 \, \mathrm{kg} \, \omega \mathrm{t} $. The correct option is
C
.

Answer

$1.6 \mathrm{~kg} \omega \mathrm{t}$

Answer

$2.0 \mathrm{~kg} \omega \mathrm{t}$

Answer

$2.4 \mathrm{~kg} \omega \mathrm{t}$

Answer

$2.8 \mathrm{~kg} \omega \mathrm{t}$

Waves

When the tension in string is increased by 3 \mathrm{~kg} \omega \mathrm{t}, the frequency of the fundamental mode increases in the ratio 2: 3. The initial tension in the string is

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