Question

The equation of a wave on a string of linear mass density \( 0.04 \) kg/m is given by:
\[ y = 0.02 \sin 2\pi \left( \frac{t}{0.04} - \frac{x}{0.50} \right) \] The tension in the string is:

• A. \( 4.0 N \)
• B. \( 12.5 N \)
• C. \( 0.5 N \)
• D. \( 6.25 N \)

Hint:

For wave motion:
- \( v = \sqrt{\frac{T}{\mu}} \).
- \( v = \frac{\omega}{k} \) relates frequency and wavelength.
- Higher tension leads to higher wave speed.

Answer 1

The Correct Option is D

Step 1: The wave velocity is given by:
\[ v = \frac{\omega}{k} \] From the given wave equation, we identify:
\[ \omega = \frac{2\pi}{0.04}, \quad k = \frac{2\pi}{0.50} \]
Step 2: The wave velocity is:
\[ v = \frac{\frac{2\pi}{0.04}}{\frac{2\pi}{0.50}} = \frac{0.50}{0.04} = 12.5 { m/s} \]
Step 3: Using \( v = \sqrt{\frac{T}{\mu}} \), solving for \( T \):
\[ T = \mu v^2 = (0.04) (12.5)^2 = 6.25 N \]