The figure represents two snaps of a travelling wave on a string of mass per unit length `mu=0.25 kg//m`. The first snap is taken at t=0 and the second is taken at `t=0.05 s`.

Determine the tension in the string.

Options

A. `25//3 N`
B. `25//7 N`
C. `5 N`
D. `25//9 N`

Answer

d

Explanation

The waves is travelling along the positivex-axis
`:. y=A sin [kx-omegat+phi]`
at `x=0, y=A sin[-omegat+phi]`
Also, at `t=0, y=A sinphi=A//2`,
`sin phi=1//2`
`impliesphi=(pi//6),(5pi//6)`,…
and at `t=0.05 s,y=A sin((-omega)/(20)+phi)=0`
or, `(-omega)/(20)+phi=0,pi,2pi`,...
For `t=0.05 s` and `x=1 m,`
`y=A sin(k-(omega)/(20)+phi)=0`
since, `lambda=2 m`
`:. pi-(omega)/(20)+phi=0,pi,2pi`,...
from Eqs. (i),(ii) and (ii), we get
`phi=pi//6`
`phi-(omega)/(20)=0`
`omega=(10pi)/(3)`
`f=(omega)/(2pi)=(5)/(3)Hz`
velocity of wave is
`V=lambdaf=(2)(5//3)=10//3 m//s`
Maximum velocity of the particle is
`V_(max) = oemga A )10 pi//3)) (10 x 10^(-3)) = pi // 30 m//s`
Tension in the string is
`T=mu V^(2)=(0.25)(10//3)^(2)=25//9N`.
the equation of the wave is
`y=10 sin [pix-(10//3)pit+(pi//6)]`