Q. The equation of a wave on a string of linear mass density $0.04\ kgm^{- 1}$ is given by $y = 0.02(m)\sin\left\lbrack 2\pi\left( \frac{t}{0.04(s)} - \frac{x}{0.50(m)} \right) \right\rbrack.$ The tension in the string is
A) $6.25\ N$
B) $4.0\ N$
C) $12.5\ N$
D) $0.5\ N$
Explanation
(a) $y = 0.02\sin\left\lbrack 2\pi\left( \frac{t}{0.04} - \frac{x}{0.50} \right) \right\rbrack$
$v = \sqrt{\frac{T}{m}} = \frac{\omega}{k} \Rightarrow \sqrt{\frac{T}{0.04}} = \frac{\frac{1}{0.04}}{\frac{1}{0.50}}$
$T = \left( \frac{0.50}{0.04} \right)^{2} \times 0.04 = (12.5)^{2} \times 0.04 = 6.25\ Newton$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.