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
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`.
C..The transverse velocity of the particle at `x=13 cm` and `t=0.015 s` will be

Options

A. `2 cm^(2)`
B. `2.5 cm^(2)`
C. `4 cm^(2)`
D. `5 cm^(2)`

Answer

b

Explanation

Shape of the pulse at `t=0`
that is a triangular pulsr.
Area of the pulse `=(1)/(2)[(4xx1)+(1xx1)]=(5)/(2)cm^(2)`