Transverse waves in stretched strings

In this section we discuss the transverse vibrations in stretched strings. We consider a uniform stretched string having a tension $T$. We take a particular section of this string which is disturbed from its mean position as shown in the figure 15.5 below. The displacement of the point $x$ on the string at time $t$ is denoted with $\xi(x,t)$. Further we assume that the disturbances are small and strictly orthogonal to the undisturbed string.

Figure 15.5: A particular section of the string

The horizontal component of the force is,

(15.19)

where $T_1$ and $T_2$ are the new tensions at points $x$ and $x+\Delta x$ respectively. Hence,

$$T_2\cos\theta_2=T_1\cos\theta_1=T.$$ (15.20)

Now coming to the vertical component of the force, we have,

$$F_y=T_2\sin\theta_2-T_1\sin\theta_1,$$ (15.21)

$$F_y=T_2\cos\theta_2\tan\theta_2-T_1\cos\theta_1\tan\theta_1.$$ (15.22)

Using the equation (15.20), we obtain,

$$F_y=T\tan\theta_2-T\tan\theta_1.$$ (15.23)

Now $\tan\theta$ at a particular point is nothing but the slope at that point of the disturbed string, so we can write,

$$F_y=T\left( {{\partial\xi({x+\Delta x},t)}\over{\partial x}}\right) -~~T\left( {{\partial\xi(x,t)}\over{\partial x}}\right),$$ (15.24)

$$F_y=T{{\partial}\over{\partial x}}\left(\xi({x+\Delta x},t)-\xi(x,t) \right),$$ (15.25)

The above force would produce the vertical acceleration in that particular section of the string. Hence,

$$F_y=\mu\Delta{x}\left({{\partial^2\xi}\over{\partial t^2}}\right),$$ (15.27)

where $\mu$ is the mass per unit length of the string. Now from equations (15.26) and (15.27) we have,

$$\left({{\partial^2\xi}\over{\partial % x^2}}\right)={{\mu}\over{T}}\left({{\partial^2\xi}\over{\partial t^2}}\right),$$ (15.28)

which can be again written as equation (15.13), that is,

$${{\partial^2\xi}\over{\partial x^2}}-{{1}\over{c_s^2}}{{\partial^2\xi}\over{\partial t^2}}=0,$$ (15.29)

where now the phase velocity, $c_s$, of the wave is equal to $\sqrt{T/\mu}$.