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For a travelling harmonic wave  $y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$, where $x$ and $y$ are in cm and $t$ in s. The phase difference between oscillatory motion of two points separated by a distance of $0.5 \text{ m}$ is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$0.8\pi \text{ rad}$

Travelling Wave Phase Difference Analysis

The equation for the travelling harmonic wave is given as:

$y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$

In this equation, $x$ and $y$ are in centimeters (cm), and $t$ is in seconds (s).

Phase Difference Calculation Steps

The phase of the wave at any point $x$ and time $t$ is the argument of the cosine function:

$\Phi(x, t) = 2\pi(10 t - 0.0080 x + 0.35)$

The phase difference $\Delta \Phi$ between two points $x_1$ and $x_2$ at the same time $t$ is calculated as:

$\Delta \Phi = \Phi(x_2, t) - \Phi(x_1, t)$

Substituting the expression for phase:

$\Delta \Phi = \left[ 2\pi(10 t - 0.0080 x_2 + 0.35) \right] - \left[ 2\pi(10 t - 0.0080 x_1 + 0.35) \right]$

Simplify the expression by canceling common terms:

$\Delta \Phi = 2\pi [ (10 t - 0.0080 x_2 + 0.35) - (10 t - 0.0080 x_1 + 0.35) ]$

$\Delta \Phi = 2\pi [ -0.0080 x_2 + 0.0080 x_1 ]$

Factor out the common term $0.0080$:

$\Delta \Phi = 2\pi \times 0.0080 (x_1 - x_2)$

Distance Unit Conversion

The distance between the two points is given as $0.5 \text{ m}$. Since the variable $x$ in the wave equation is in centimeters, we must convert the distance to centimeters:

$\Delta x = x_1 - x_2 = 0.5 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 50 \text{ cm}$

Final Phase Difference Calculation

Now substitute the distance $\Delta x$ into the phase difference formula. We consider the magnitude of the phase difference:

$|\Delta \Phi| = |2\pi \times 0.0080 \times \Delta x|$

$|\Delta \Phi| = 2\pi \times 0.0080 \times 50$

Calculate the product:

$|\Delta \Phi| = 2\pi \times 0.4$

$|\Delta \Phi| = 0.8\pi \text{ rad}$

The phase difference between the two points separated by $0.5 \text{ m}$ is $0.8\pi$ radians.

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