Question 25: Waves

Question

A string vibrates with a frequency of $$200 \mathrm{~Hz}$$. When its length is doubled and tension is altered, it begins to vibrate with a frequency of $$300 \mathrm{~Hz}$$. The ratio of the new tension to the original tension is

Accepted Answer

The frequency of a vibrating string can be expressed by the formula:
$$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$
Where:
$$f$$ is the frequency of the string,
$$L$$ is the length of the string,
$$T$$ is the tension in the string, and
$$\mu$$ is the linear mass density of the string.
Initially, the frequency of the string is $$200 \mathrm{~Hz}$$ when the length is $$L$$ and the tension is $$T_1$$. The formula becomes:
$$ 200 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} $$
When the length of the string is doubled to $$2L$$ and the tension is altered to $$T_2$$, the frequency becomes $$300 \mathrm{~Hz}$$. The formula then changes to:
$$ 300 = \frac{1}{4L} \sqrt{\frac{T_2}{\mu}} $$
We can rearrange each equation to express the square root of tension over linear mass density:
$$ \sqrt{\frac{T_1}{\mu}} = 400 L $$
$$ \sqrt{\frac{T_2}{\mu}} = 1200 L $$
Squaring both sides of each equation gives:
$$ \frac{T_1}{\mu} = (400 L)^2 $$
$$ \frac{T_2}{\mu} = (1200 L)^2 $$
Take the ratio $$ \frac{T_2}{T_1} $$:
$$ \frac{T_2}{T_1} = \frac{(1200 L)^2}{(400 L)^2} $$
$$ \frac{T_2}{T_1} = \left(\frac{1200}{400}\right)^2 $$
$$ \frac{T_2}{T_1} = (3)^2 $$
$$ \frac{T_2}{T_1} = 9 $$
Thus, the ratio of the new tension $$T_2$$ to the original tension $$T_1$$ is 9:1, indicating that the new tension is 9 times greater than the original tension.
The correct answer is
Option A: 9 : 1
.

Answer

9 : 1

Answer

1 : 9

Answer

3 : 1

Answer

1 : 3

Waves

A string vibrates with a frequency of $200 \mathrm{~Hz}. When its length is doubled and tension is altered, it begins to vibrate with a frequency of 300 \mathrm{~Hz}$. The ratio of the new tension to the original tension is

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