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Q15.12(i)(b):
For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (b) phase? Explain your answers.

Solution :

1. Theoretical Foundation of Phase in Traveling Waves

The displacement of a transverse wave traveling along a string is defined by the mathematical function: $y(x, t) = A \sin(kx - \omega t + \phi)$ where:

  • $A$ is the amplitude.
  • $k$ is the angular wave number ($\frac{2\pi}{\lambda}$).
  • $\omega$ is the angular frequency ($2\pi f$).
  • $(kx - \omega t + \phi)$ represents the phase ($\Phi$) of the wave at position $x$ and time $t$.

2. Analysis of Phase Variation

To determine if all points on the string oscillate with the same phase, we examine the phase difference ($\Delta\Phi$) between two arbitrary points on the string, $x_1$ and $x_2$, at any fixed instant of time $t$.

The phase at $x_1$ is $\Phi_1 = kx_1 - \omega t + \phi$.
The phase at $x_2$ is $\Phi_2 = kx_2 - \omega t + \phi$.

The phase difference is given by: $\Delta\Phi = \Phi_2 - \Phi_1 = (kx_2 - \omega t + \phi) - (kx_1 - \omega t + \phi)$ $\Delta\Phi = k(x_2 - x_1) = k\Delta x$

3. Mathematical Justification

Since $k = \frac{2\pi}{\lambda}$, the phase difference depends explicitly on the spatial separation $\Delta x$ between the points: $\Delta\Phi = \frac{2\pi}{\lambda} \Delta x$

For all points to oscillate with the same phase, the condition $\Delta\Phi = 0$ (or an integer multiple of $2\pi$) must hold for any arbitrary distance $\Delta x$. However, for a traveling wave where $k \neq 0$, the phase difference is a linear function of $x$. Therefore, points separated by any distance that is not an integer multiple of the wavelength $\lambda$ will necessarily have different phases.

4. Visual Representation of Phase Shift

Point A Point B At any instant t, position A and B have different vertical displacements.

Conclusion

In a traveling wave, the phase is a function of the spatial coordinate $x$. Points separated by a distance $\Delta x$ exhibit a phase difference of $\frac{2\pi}{\lambda} \Delta x$. Because this value is non-zero for arbitrary points, the points on the string do not oscillate in phase. They oscillate with the same frequency $\omega$, but with a phase lag or lead relative to one another depending on their spatial positions.

Final Solution: No, all points on the string do not oscillate with the same phase. The phase of each point depends on its position $x$ along the string, specifically defined by the term $(kx - \omega t + \phi)$.


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