A pulse is started at a time `t=0` along the `+x`direction on a long, taut string. The shape of the pulse at `t=0` is given by function `f(x)` with ` {((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}`
`0`, otherwise
`{((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}`
`0`, otherwise
here `f` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`.
The verticle displacement of the particle of the string at `x=7 cm` and `t=0.01 s` will be

Options

A. `0.75 cm`
B. `0.5 cm`
C. `0.25 cm`
D. `zero

Answer

c

Explanation

`v=sqrt((T)/(mu))=10 m//s`
solution of the wave equation that gives displacement of any piece of the string at any time
`{((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}`
`0`, otherwise
Using `v=1000 cm//s,t=0.01s`
`vt=10 cm`
as `(vt-4)lt(x=7 cm)ltvt`
`y=(1)/(4)(7-10)+1+(1)/(4)cm=0.25cm`