To determine the speed of transverse wave in a stretched string, consider a pulse on a stretched string travelling in the positive x-direction with constant speed $v$ as in Figure 1. The pulse is travelling towards the right and in the pulse's frame of reference you see that the string is moving towards the left; if you consider a particular element of the string on the right of the pulse, you'll see the element follows the curve of the pulse and moves towards the left of the pulse or in negative x-direction.

In Figure 1 we have considered an element of the string of length $ds$ on the right of the pulse which makes an approximate arc of the same length $ds$ when it reaches the top of the pulse.

The same magnitude of tension force $F$ acts on both ends of the arc. The arc has extremely small length $ds$, so we can take it as circular in a circle of radius $R$, and therefore $ds = R(2\theta)$. If $\mu$ is the mass per unit length of the string also called linear density, the mass of the element of the string of length $ds$ is $dm = \mu ds = \mu R (2\theta)$.

The horizontal components of tension $F$ are equal and opposite and cancel each other while the vertical components add together to provide the centripetal acceleration to the length element $ds$ as it follows the curve of the pulse. Therefore, the the sum of the vertical components of tension is $F\sin \theta +F\sin \theta =2F\sin \theta $ which is equal to the centripetal force on the arc:

\[2F\sin \theta =\frac{dm\,{{v}^{2}}}{R}\]

Since the angle $\theta$ is considered to be small enough so that we can replace $\sin \theta$ in the above equation by simply $\theta$. The small angle $\theta$ means that the wave pulse is almost horizontal so that the tension $F$ in the pulse is the same as the tension in the string in undisturbed condition. Therefore, the tension $F$ equals the tension in the string in undisturbed condition.

\[\begin{align*} 2F\theta &=\frac{\mu R(2\theta ){{v}^{2}}}{R} \\ \text{or,}\quad {{v}^{}}&=\sqrt{\frac{F}{\mu }} \tag{1} \label{1} \end{align*}\]

The above equation Eq. \eqref{1} gives the wave speed of a transverse wave along a stretched string. As you can see the wave speed is directly proportional to the square root of the tension and inversely proportional to the square root of the linear density.

As the tension increases the wave speed increases and as the linear density increases the wave speed decreases which seems reasonable. As the tension increases, it makes the disturbance travel much faster. Linear density represents the inertia which resists the wave and decreases the speed when it increases.

We didn't make any special assumptions about the shape of the pulse while deriving the Eq. \eqref{1}, so the transverse wave pulse need not have a simple harmonic nature but can have any shape and still the wave speed is given by Eq. \eqref{1}.

The above derivation of the speed of transverse wave along a string is useful but we also discuss another way to find the speed of the same transverse wave. You may consider reading both or one whichever you like.

## Speed of Transverse Wave Alternative Method

We again consider a string stretched under tension of magnitude $F_\text{x}$ at both of its ends along the x-axis of our coordinate system. As the left end of the string is moved vertically upward as in Figure 2 by applying force $F_\text{y}$, the successive particles from the left end of the string start to come in motion.

The force $F_\text{y}$ is applied in such a way that all particles in motion will have the same vertical velocity including the left end of the string. The movable point $p$ separates the string in motion and the string at rest.

As the left end of the string is moved upward, the movable point $p$ moves towards the right end with the speed equal to the wave speed $v$. Since all particles in motion move upward with the same upward velocity $v_\text{y}$, the wave speed $v$ is constant.

The impulse due to the force $F_\text{y}$ in time $t$ is $F_\text{y} t$ which is equal to the change in momentum of the system. The initial momentum before applying the force $F_\text{y}$ is zero but the final momentum in time $t$ is $m v_\text{y}$ where $m$ is the mass of the segment of the string in motion. Thus:

\[{{F}_{y}}t=m{{v}_{y}} \tag{2} \label{2}\]

The movable point moves a distance $vt$ in time $t$ and similarly the left end of the string moves upward a distance of $v_\text{y} t$ in the same time.

If $\mu$ is the linear density or mass per unit length, the mass of the string segment in motion in time $t$ is $m = \mu vt$. As the velocity $v_\text{y}$ is constant, the change in momentum is due to the change in mass of the string segment in motion.

In Figure 2 the horizontal component of tension on the left end remains the same which is $F_\text{x}$ but the overall tension on the left end is the vector sum of the two perpendicular forces $F_\text{x}$ and $F_\text{y}$ which is greater than the tension on the right end and the string is somewhat more stretched in the moving segment of the string. The two shaded triangles in Figure 2 are similar and hence you can find,

\[\begin{align*} \frac{{{F}_{y}}}{{{F}_{x}}}&=\frac{{{v}_{y}}t}{vt} \\ \text{or,}\quad {{F}_{y}}&=\frac{{{v}_{y}}{{F}_{x}}}{v} \tag{3} \label{3} \end{align*}\]

You can substitute the values of $F_\text{y}$ form Eq. \eqref{3} and $m = \mu vt$ into Eq. \eqref{2}, we can find,

\[\begin{align*} \frac{{{v_y}{F_x}}}{v}t &= \mu vt{\kern 1pt} {v_y}\\ {\rm{or,}}\quad v &= \sqrt {\frac{{{F_x}}}{\mu }} = \sqrt {\frac{F}{\mu }} \tag{4} \label{4} \end{align*}\]

Eq. \eqref{4} is the same expression of wave speed of a transverse wave along a string we already found in Eq. \eqref{1}. But here we made a special assumption of constant upward speed $v_\text{y}$ but it does not appear in Eq. \eqref{4} which again verifies that the speed of wave given by Eqs. \eqref{1} or \eqref{4} is the same for the wave pulse of any shape.

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