Physics - Wave on String Question with Solution | TestHub

The total energy carried by the wave is the initial potential energy stored, as energy is conserved. Define the origin to be at the left end of the string. The equations of the left and right segments are $\mathrm{y}=\frac{\mathrm{h}}{\mathrm{L}}\mathrm{x}$ and $\mathrm{y}=\mathrm{h}-\frac{\mathrm{h}}{\mathrm{L}}\mathrm{x}$ respectively. The magnitude of the gradients is $\frac{\mathrm{h}}{\mathrm{L}}$. Therefore, the stored potential energy is

$$\begin{array}{rlrl}& \mathrm{E}={\int }_0^{2\mathrm{L}}\frac{1}{2}\mathrm{F}& ={\int }_0^{2\mathrm{L}}\frac{{\mathrm{Fh}}^2}{2{\mathrm{L}}^2}\mathrm{dx}& =\frac{{\mathrm{Fh}}^2}{\mathrm{L}}\end{array}$$

To determine the period of the resultant wave, observe that the resultant wave can be decomposed into the superposition into the superposition of two smaller triangular waves (scaled down vertically by a factor of $\frac{1}{2}$ and extended beyond the walls, as shown in fig.) traveling in opposite directions at speed $\mathrm{v}=\sqrt{\frac{\mathrm{F}}{\mu }}$. The actual wave is composed of two of the above "sub-waves" traveling in opposite directions. After the two waves have "covered" 4 L distance each (this the wavelength), the string returns to its original state. Therefore, the period is

$$\mathrm{T}=\frac{4\mathrm{L}}{\mathrm{v}}=\frac{4\mathrm{L}\sqrt{\mu }}{\sqrt{\mathrm{F}}}$$

At time $\mathrm{t}=\frac{\mathrm{T}}{8}$, the two waves would have traveled distance $\frac{\mathrm{L}}{2}$ in opposite directions - their positions at this juncture are depicted below. Their superposition yields the following trapezoid.