Question

A sonometer string vibrates with a frequency of \(400\,Hz\). When the length of the string is halved and the tension is altered, it begins to vibrate with a frequency of \(200\,Hz\). The ratio of the new tension to the original tension in the string is:

• A. \(4:1\)
• B. \(16:1\)
• C. \(1:4\)
• D. \(1:16\)

Hint:

For stretched string, \(f \propto \frac{\sqrt{T}}{l}\). First use length change, then calculate tension ratio.

Answer 1

The Correct Option is D

Step 1: Frequency of vibrating string.
For a sonometer string, frequency is given by:
\[ f = \frac{1}{2l}\sqrt{\frac{T}{\mu}} \]

Step 2: Write proportionality relation.
For same string, \(\mu\) remains constant.
\[ f \propto \frac{\sqrt{T}}{l} \]

Step 3: Write ratio of frequencies.
\[ \frac{f_2}{f_1} = \frac{\sqrt{T_2}/l_2}{\sqrt{T_1}/l_1} \]
\[ \frac{f_2}{f_1} = \frac{l_1}{l_2}\sqrt{\frac{T_2}{T_1}} \]

Step 4: Substitute given values.
Given:
\[ f_1 = 400\,Hz,\quad f_2 = 200\,Hz,\quad l_2 = \frac{l_1}{2} \]
\[ \frac{200}{400} = \frac{l_1}{l_1/2}\sqrt{\frac{T_2}{T_1}} \]

Step 5: Simplify equation.
\[ \frac{1}{2} = 2\sqrt{\frac{T_2}{T_1}} \]
\[ \sqrt{\frac{T_2}{T_1}} = \frac{1}{4} \]

Step 6: Square both sides.
\[ \frac{T_2}{T_1} = \frac{1}{16} \]

Step 7: Final conclusion.
\[ \boxed{1:16} \] Hence, correct answer is option (D).