Question

A transverse wave described by \[y = \left( 0 \cdot 02  m \right)  \sin  \left( 1 \cdot 0  m^{- 1} \right)  x + \left( 30  s^{- 1} \right)t\] propagates on a stretched string having a linear mass density of \[1 \cdot 2 \times  {10}^{- 4}   kg   m^{- 1}\] the tension in the string.

Answer 1

Given,
Wave equation,
\[y = \left( 0 \cdot 02  m \right)\sin\left( 1 \cdot 0  m^{- 1} \right)x + \left( 30  s^{- 1} \right)t\]
Let:
\[\text{ Mass  per  unit  length,   m = 1 . 2 \times  {10}^{- 4}   kg/m }\]
From  the  wave  equation,   we  have:
\[k = 1   m^{- 1}  = \frac{2\pi}{\lambda}\]
And,
\[\omega = 30   s^{- 1}  = 2\pi f\]
Velocity of the wave in the stretched string is given by
\[\nu = \lambda f = \frac{\omega}{k} = \frac{30}{1}\]
\[ \Rightarrow v = 30  m/s\]
We  know:
\[v = \sqrt{\frac{T}{m}}\]
\[ \Rightarrow 30   = \sqrt{\left( \frac{T}{1 . 2 \times {10}^{- 4}} \right)}\]
\[ \Rightarrow T = 108 \times  {10}^{- 3}  = 0 . 108  N\]
So, the tension in the string is 0.108 N.