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Grade 12Wave Motion

The stading wave on a 5 m long string clamped at both ends is represented by the equation y=4sin6.28xcos6.28t, where y is in cm. The phase difference between two points at x=1.51m and x=2.75 m is

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11 Years agoGrade 12
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ApprovedApproved Tutor Answer1 Year ago

To determine the phase difference between two points on a standing wave represented by the equation \( y = 4 \sin(6.28x) \cos(6.28t) \), we first need to understand how the wave function works. The equation describes a standing wave formed by the superposition of two traveling waves moving in opposite directions. The key components of the equation are the sine and cosine functions, which indicate the spatial and temporal characteristics of the wave.

Understanding the Wave Equation

The equation can be broken down into two parts:

  • Spatial Component: \( \sin(6.28x) \)
  • Temporal Component: \( \cos(6.28t) \)

Here, \( 6.28 \) is the angular wave number \( k \), which relates to the wavelength \( \lambda \) of the wave. The wavelength can be calculated using the formula \( k = \frac{2\pi}{\lambda} \). In this case, \( \lambda \) can be derived as follows:

Since \( k = 6.28 \), we have:

\( \lambda = \frac{2\pi}{6.28} \approx 1 \, \text{m} \)

Calculating the Phase at Specific Points

Next, we need to find the phase of the wave at the two given positions, \( x = 1.51 \, \text{m} \) and \( x = 2.75 \, \text{m} \). The phase \( \phi \) at any point \( x \) is given by:

\( \phi = 6.28x \)

Now, let's calculate the phase at both positions:

  • For \( x = 1.51 \, \text{m} \):
  • \( \phi_1 = 6.28 \times 1.51 \approx 9.47 \, \text{radians} \)

  • For \( x = 2.75 \, \text{m} \):
  • \( \phi_2 = 6.28 \times 2.75 \approx 17.29 \, \text{radians} \)

Finding the Phase Difference

The phase difference \( \Delta \phi \) between the two points is simply the absolute difference between their phases:

\( \Delta \phi = |\phi_2 - \phi_1| = |17.29 - 9.47| \approx 7.82 \, \text{radians} \)

Final Thoughts

Thus, the phase difference between the two points at \( x = 1.51 \, \text{m} \) and \( x = 2.75 \, \text{m} \) is approximately \( 7.82 \, \text{radians} \). This phase difference indicates how much one point is ahead or behind the other in the oscillation of the wave at a given moment in time. Understanding these concepts is crucial in wave mechanics, as they help us analyze and predict wave behavior in various physical contexts.