If the amplitudes of the two waves are 3 units and 1 unit respectively, show by the principle of superposition that the ratio of the amplitudes of the stationary wave at an antinode and node respectively 2:1

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That would be the ratio of the sum of the two amplitudes (when they are in phase or, as they say, "add coherently") to the difference (when they are out of phase or "add destructively")

(3+1)/(3-1) = ?

The power ratio would be the square of that.

To demonstrate the ratio of amplitudes of the stationary wave at an antinode and a node, we need to consider the principle of superposition.

In the principle of superposition, two waves of equal frequency and amplitude traveling in opposite directions are combined to create a stationary wave pattern. The combined wave is the sum of the two individual waves at each point in space and time.

Let's suppose we have two waves with amplitudes 3 units and 1 unit respectively.

When these waves combine, we get the following equation for the combination of the waves:

\[ y = A_1 \sin(kx - \omega t) + A_2 \sin(kx + \omega t) \]

where,
- \(A_1\) and \(A_2\) are the amplitudes of the waves,
- \(k\) is the wave number,
- \(x\) is the position,
- \(\omega\) is the angular frequency, and
- \(t\) is the time.

If we substitute the given amplitudes into the equation, we get:

\[ y = 3 \sin(kx - \omega t) + 1 \sin(kx + \omega t) \]

To find the amplitude at an antinode, we need to find a point where the amplitude is maximum. This occurs when the two terms in the equation have the same sign, either both positive or both negative. In this case, when both the terms are positive, the amplitude is 3 + 1 = 4 units. Therefore, the amplitude at an antinode is 4 units.

On the other hand, to find the amplitude at a node, we need to find a point where the amplitude is minimum or zero. This occurs when the two terms in the equation have opposite signs. In this case, when one term is positive (3) and the other is negative (1), the amplitude is 3 - 1 = 2 units. Therefore, the amplitude at a node is 2 units.

Hence, the ratio of the amplitudes of the stationary wave at an antinode and a node is 4:2, which simplifies to 2:1.

To demonstrate the principle of superposition and determine the ratio of amplitudes of a stationary wave at an antinode and a node, we need to consider two waves of different amplitudes.

Let's suppose we have two waves with amplitudes 3 units (wave A) and 1 unit (wave B) respectively. When these two waves superpose (overlap) and interfere, a stationary wave is produced.

In a stationary wave, some points experience maximum displacement (called antinodes) while others experience zero displacement (called nodes). We want to calculate the ratio of the amplitudes of the stationary wave at an antinode and a node.

The principle of superposition states that when two waves superpose, the displacement of the resulting wave is the algebraic sum of the displacements of the individual waves at that point.

At an antinode, the displacements of wave A and wave B are in the same direction, so we add the amplitudes:
Displacement at antinode = Amplitude of wave A + Amplitude of wave B = 3 + 1 = 4 units

At a node, the displacements of wave A and wave B are in opposite directions, so we subtract the amplitudes:
Displacement at node = Amplitude of wave A - Amplitude of wave B = 3 - 1 = 2 units

Therefore, the ratio of the amplitudes of the stationary wave at an antinode and a node is 4:2, which simplifies to 2:1.