Question 4.7.7: A cord of length P and mass per unit length μ is under tensi......

For a general solution of the transverse vibration of the cord, one can use

\operatorname{y}(x,t)=\left[A\sin\left({\frac{\omega x}{c}}\right)+B\cos\left({\frac{\omega x}{c}}\right)\right]\sin\omega t,

where c={\sqrt{T/\mu}} is the velocity of wave propagation along the cord.

At x = 0, y(0, t) = 0, because it is fixed at this end. This leads to B = 0.

At x = ℓ, y(ℓ, t) = Y.

With reference to the FBD (Figure 4E7b) and by Newton’s law of motion applied at x = ℓ to the spring-mass system, one has

m{\frac{\partial^{2}Y}{\partial t^{2}}}=-k Y-T\sin\theta|_{x=l}.

Since θ is small, \sin\theta\approx\theta.\;\mathrm{Also,}\theta={\frac{\partial y}{\partial x}}. Thus, the last equation can be written as

m{\frac{\partial^{2}Y}{\partial t^{2}}}+k Y=-T{\frac{\partial y}{\partial x}}{\biggl|}_{x=\ell}.              (i)

As small oscillation or linear vibration is assumed, the method of separable variables can be applied. Thus, one can write

\operatorname{y}(x,t)=W(x)\sin\omega t,\ \ {\mathrm{where}}\ W(x)=A\sin\left({\frac{\omega x}{c}}\right).

{\frac{\partial y}{\partial x}}{\Big|}_{x=\ell}=A{\Big(}{\frac{\omega}{c}}{\Big)}\cos\left({\frac{\omega \ell}{c}}\right)\sin\omega t.

Substituting this equation into Equation (i),

m{\frac{{\partial}^{2}Y}{\partial t^{2}}}+k Y=-T\left({\frac{\omega}{c}}\right)A\,\cos\left({\frac{\omega \ell}{c}}\right)\sin\omega t\ \mathrm{~~or}

-m\omega^{2}W(\ell)\sin\omega t+k W(\ell)\sin\omega t=-T\Bigl({\frac{\omega}{c}}\Bigr)A\cos\left({\frac{\omega \ell}{c}}\right)\sin\omega t.

Simplifying,

W({\ell})=\frac{-{T}\left(\frac{\omega}{c}\right)A\cos\left(\frac{\omega{\ell}}{c}\right)}{k-m\omega^{2}}.        (ii)

Equation (ii) is required for the determination of natural frequencies of the given cord-springmass system.