4.0 Speed of a transverse wave on a string

  Motion of Waves

Consider a transverse wave produced in a tensioned string of linear mass density $\mu $. Consider a small segment of the wave of length $\Delta l$, forming an arc of a circle of radius $R$.

A force equal in magnitude to the tension $T$ pull tangentially on this segment at each end.

Let an observer be at the centre $O$ of the pulse which moves alongwith the pulse towards right.

Consider a segement of the string of length $\Delta l$ which is in a circular path of raidus $R$ moving with speed $v$ as shown in the figure.

Let $\Delta m$ be the mass of the segment of string. Mathematically, $$\Delta m = \mu \Delta l$$ Also, $$\Delta l = R(2\theta )$$ So, $$\Delta m = 2R\theta \mu \quad ...(i)$$

Therefore, the required required centrepetal force is provided by the component of tension ($T$). From the FBD we can write, $$2T\sin \theta = \frac{{\Delta m{v^2}}}{R}\quad ...(ii)$$ From equation $(i)$ and $(ii)$ we get, $$2T\sin \theta = \frac{{2R\theta \mu {v^2}}}{R}$$ As $\theta $ is very small. So, $$\sin \theta \approx \theta $$ Therefore, $$\begin{equation} \begin{aligned} 2T\theta = \frac{{2R\theta \mu {v^2}}}{R} \\ v = \sqrt {\frac{T}{\mu }} \\\end{aligned} \end{equation} $$