Question

A transverse displacement of vibrating string is \( y = 0.06 \sin\left(\frac{2 \pi}{3} x\right) \cos(120 \pi t) \). If the mass per unit length of a string is \( 4 \times 10^{-2} \) kg/m, then the tension in the string will be

• A. 1296 N
• B. 1000 N
• C. 720 N
• D. 500 N

Hint:

The tension in the string can be calculated using the wave velocity and the mass per unit length. The wave velocity on the string is related to the tension and linear mass density.

Answer 1

The Correct Option is C

Step 1: Formula for the wave velocity.
The wave on the string is described by the equation:
\[ y = A \sin(kx) \cos(\omega t), \]
where \( k \) is the wave number and \( \omega \) is the angular frequency.
For the given string, \( k = \frac{2\pi}{3} \) and \( \omega = 120\pi \).
The velocity of the wave \( v \) on a string is given by:
\[ v = \frac{\omega}{k}. \]
Substituting the given values of \( \omega \) and \( k \):
\[ v = \frac{120\pi}{\frac{2\pi}{3}} = 180 \, \text{m/s}. \]

Step 2: Relationship between velocity, tension, and linear mass density.
The velocity of a wave on a string is also related to the tension \( T \) and the mass per unit length \( \mu \) by the formula:
\[ v = \sqrt{\frac{T}{\mu}}. \]
Rearranging for the tension \( T \):
\[ T = \mu v^2. \]

Step 3: Substituting values to find the tension.
We are given \( \mu = 4 \times 10^{-2} \, \text{kg/m} \) and \( v = 180 \, \text{m/s} \), so:
\[ T = (4 \times 10^{-2}) \times (180)^2 = 4 \times 10^{-2} \times 32400 = 720 \, \text{N}. \]
Final Answer:
Thus, the tension in the string is:
\[ \boxed{720 \, \text{N}}. \]