Q. A transverse wave propagating on a stretched string of liner density $3 \times 10^{- 4}\ kg -$m^(-1) is represented by the equation y=0.2 sin(1.5x+60t) where x is in meter and t is in second. The tension in the string (in newton) is
A) 0.24
B) 0.48
C) 1.20
D) 1.80
Explanation
(b) Equation of wave y=0.2 sin (1.5x+60t)
Comparing with standard equation
y=A sin (kx+ωt)
k=1.5, ω=60
$\therefore\ \ velocity\ of\ wave\ v = \frac{\omega}{k} = \frac{60}{1.5} = 40ms^{- 1}$
Velocity of wave on a stretched string
$v = \sqrt{\frac{T}{m}}$
Where m-linear density
T= tension in the string
So, $T = v^{2}m$
$= (40)^{2} \times 3 \times 10^{- 4} = 0.48$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.