Question

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

• A. 9:1
• B. 1:9
• C. 3:1
• D. 1:3

Hint:

The frequency of a vibrating string is proportional to the square root of the tension and inversely proportional to the square root of the length. A change in length or tension will significantly affect the frequency.

Answer 1

The Correct Option is A

To determine the ratio of the new tension to the original tension for the vibrating string, we should consider the fundamental frequency formula for a stretched string:

\( f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}} \)

Where:

• \( f \) is the frequency of vibration.
• \( L \) is the length of the string.
• \( T \) is the tension in the string.
• \( \mu \) is the linear mass density of the string, assumed constant in both cases.

Initially, the frequency \( f_1 \) is 200 Hz, and the length is \( L \).

When the string length is doubled to \( 2L \) and the tension changes, the frequency becomes \( f_2 = 300 \) Hz.

Using the formula for frequency in both scenarios, we have:

\( f_1 = \dfrac{1}{2L}\sqrt{\dfrac{T_1}{\mu}} = 200 \)

\( f_2 = \dfrac{1}{4L}\sqrt{\dfrac{T_2}{\mu}} = 300 \)

Divide the second equation by the first equation:

\( \dfrac{f_2}{f_1} = \dfrac{\dfrac{1}{4L}\sqrt{\dfrac{T_2}{\mu}}}{\dfrac{1}{2L}\sqrt{\dfrac{T_1}{\mu}}} \)

After simplification:

\( \dfrac{300}{200} = \dfrac{1}{2}\sqrt{\dfrac{T_2}{T_1}} \)

By solving the frequency ratio, we find:

\( \dfrac{3}{2} = \dfrac{1}{2}\sqrt{\dfrac{T_2}{T_1}} \)

Now, solve for \( \dfrac{T_2}{T_1} \):

\( \sqrt{\dfrac{T_2}{T_1}} = 3 \)

Squaring both sides yields:

\( \dfrac{T_2}{T_1} = 9 \)

Hence, the ratio of the new tension to the original tension is 9:1.

Answer 2

To solve this problem, we need to understand the relationship between the frequency of a vibrating string, its length, and the tension applied to it. The frequency \( f \) of a vibrating string is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density of the string. Assuming the linear mass density remains constant, the frequency is directly proportional to the square root of the tension and inversely proportional to the length of the string.

Initially, the frequency \( f_1 = 200 \, \text{Hz} \) with length \( L \) and tension \( T_1 \). When the length is doubled, the new frequency \( f_2 = 300 \, \text{Hz} \), length \( L_2 = 2L \), and tension \( T_2 \). We can set up the ratio of the two frequencies using:

\[ \frac{f_2}{f_1} = \frac{\frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}} \]

Substitute \( L_2 = 2L \) and simplify:

\[ \frac{300}{200} = \frac{\frac{1}{4L} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}} = \frac{1}{2} \times \sqrt{\frac{T_2}{T_1}} \]

\[ \frac{3}{2} = \frac{1}{2} \times \sqrt{\frac{T_2}{T_1}} \]

Solving for the tension ratio:

\[ \frac{3}{2} \times 2 = \sqrt{\frac{T_2}{T_1}} \]

\[ 3 = \sqrt{\frac{T_2}{T_1}} \]

Squaring both sides, we find:

\[ 9 = \frac{T_2}{T_1} \]

Thus, the ratio of the new tension to the original tension is \( 9:1 \)