Question #171473

A tightly stretched string of length ‘l’ has its ends fastened at x=0, x=l. The midpoint

of the string is then taken to height ‘h’ and then released from rest in that

position. Find the initial displacement y(x,0) of the string.


1
Expert's answer
2021-03-16T17:06:30-0400

When string is released, then it will vibrate. Let the equation which represents the displacement is given by, y(x,t)=Asin(ωt+kx+ϕ)y(x,t)= Asin(\omega t +kx+\phi)

where omega is angular velocity, k is wave number, phi is initial phase.


According to the question, at end points x=0 and x=l, displacement is zero and maximum displacement is at mid point that is h.

So, A=hA=h


Now, since string is held at end points so it represents half wavelength of the wave. Then k=2πλ,λ=l/2    k=4πlk=\frac{2\pi}{\lambda}, \lambda =l/2 \implies k=\frac{4\pi}{l}

Since we are asked at t=0, so equation will look like y(x,0)=hsin(kx+ϕ)y(x,0)= hsin(kx+\phi)


For phase, at x=0 and x=l, displacement is zero, so

0=hsin(ϕ)0= hsin(\phi) and. 0=hsin(4πll+ϕ)=hsin(4π+ϕ)0 = hsin(\frac{4π}{l}l+\phi)= hsin(4π+\phi)


Solving these, we get ϕ=0\phi = 0

Hence final equation will be, y(x,0)=hsin(4πlx)y(x,0)= hsin( \frac{4π}{l} x)




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