InterviewSolution
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InterviewSolution
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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 beA. `0.75 cm`B. `0.5 cm`C. `0.25 cm`D. `zero |
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Answer» Correct Answer - c `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` |
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