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.


Expert's answer

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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