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Question

Physics Question on Waves

A uniform string of length 20m20 \,m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is : (take g=10ms2g\, = \,10\,ms^{-2})

A

2π2s2 \pi \sqrt{2} s

B

2s2s

C

22s2 \sqrt{2} s

D

2s\sqrt{2} s

Answer

22s2 \sqrt{2} s

Explanation

Solution

dydt=gyρAμ\frac{dy}{dt} = \sqrt{\frac{gy \rho A}{\mu}}
dydt=gy\frac{dy}{dt} = \int \sqrt{gy}
dyy=gdt\int \frac{dy}{\sqrt{y}} = \sqrt{g} dt
y12+112+10l=gt0t\frac{y^{- \frac{1}{2}+1}}{-\frac{1}{2}+1}|^{l}_{0} = \sqrt{g } t |^{t}_{0}
t=22010=22sect = 2 \sqrt{\frac{20}{10}} = 2 \sqrt{2} \sec