Question

A string of length \( L \) is stretched by \( \frac{L}{20} \) and the speed of transverse waves along it is \( v \). The speed of the wave when it is stretched by \( \frac{L}{10} \) will be:

• A. \( 2v \)
• B. \( \frac{v}{\sqrt{2}} \)
• C. \( v\sqrt{2} \)
• D. \( 4v \)

Hint:

For waves in a string, remember that the speed is proportional to the square root of the tension, which itself is proportional to the elongation (according to Hooke's law).

Answer 1

The Correct Option is C

The speed of the wave in a string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension in the string and \( \mu \) is the mass per unit length. According to Hooke's law, the tension \( T \) is directly proportional to the elongation of the string. So, if the length of the string increases, the tension increases in proportion to the elongation. Given that the elongation is increased from \( \frac{L}{20} \) to \( \frac{L}{10} \), the tension increases by a factor of \( \frac{L}{10} \div \frac{L}{20} = 2 \). Since the speed of the wave is proportional to the square root of the tension, the speed of the wave increases by a factor of \( \sqrt{2} \). Thus, the speed of the wave when the string is stretched by \( \frac{L}{10} \) will be \( v\sqrt{2} \). Therefore, the correct answer is option (3), \( v\sqrt{2} \).

Answer 2

Step 1: Understanding the relationship
The speed of transverse waves on a stretched string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the mass per unit length.

Step 2: Relation between tension and extension
Tension \( T \) in the string is proportional to the extension if the string follows Hooke's law:
\[ T \propto \text{extension} \]

Step 3: Given data
- Initial extension = \( \frac{L}{20} \), corresponding wave speed = \( v \)
- New extension = \( \frac{L}{10} \) (which is twice the initial extension)

Step 4: Calculate the new speed
Since tension \( T \) doubles (because extension doubles), new tension \( T' = 2T \).
Then the new speed \( v' \) is:
\[ v' = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{2T}{\mu}} = \sqrt{2} \times \sqrt{\frac{T}{\mu}} = v \sqrt{2} \]

Step 5: Conclusion
Therefore, when the string is stretched by \( \frac{L}{10} \), the speed of the transverse wave becomes \( v \sqrt{2} \).