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Find the resultant of the superposition of two harmonic waves in the form ๐ธ = ๐ธ0 ๐‘๐‘œ๐‘  ๐‘๐‘œ๐‘  (๐›ผ โˆ’ ๐œ”๐‘ก) ,with amplitudes of 3 and 4 units and phases of ๐œ‹6 and ๐œ‹2 respectively. Both waves have a period of 1 s

Question

Find the resultant of the superposition of two harmonic waves in the form ๐ธ = ๐ธ0 ๐‘๐‘œ๐‘  ๐‘๐‘œ๐‘  (๐›ผ โˆ’ ๐œ”๐‘ก) ,with amplitudes of 3 and 4 units and phases of ๐œ‹6 and ๐œ‹2 respectively. Both waves have a period of 1 s

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Solution

To find the resultant of the superposition of two harmonic waves, we need to add the individual wave equations together.

The given wave equations are ๐ธ1 = ๐ธ0๐‘๐‘œ๐‘ (๐›ผ1 โˆ’ ๐œ”๐‘ก) and ๐ธ2 = ๐ธ0๐‘๐‘œ๐‘ (๐›ผ2 โˆ’ ๐œ”๐‘ก), where ๐ธ0 is the amplitude, ๐›ผ is the phase, and ๐œ” is the angular frequency.

For the first wave, ๐ธ1, the amplitude is 3 units and the phase is ๐œ‹/6. The angular frequency, ๐œ”, can be calculated using the formula ๐œ” = 2๐œ‹/T, where T is the period. In this case, the period is 1 second, so ๐œ” = 2๐œ‹/1 = 2๐œ‹ rad/s.

Substituting the values into the wave equation, ๐ธ1 = 3๐‘๐‘œ๐‘ (๐›ผ1 โˆ’ 2๐œ‹๐‘ก).

For the second wave, ๐ธ2, the amplitude is 4 units and the phase is ๐œ‹/2. Using the same formula for ๐œ”, we find ๐œ” = 2๐œ‹ rad/s.

Substituting the values into the wave equation, ๐ธ2 = 4๐‘๐‘œ๐‘ (๐›ผ2 โˆ’ 2๐œ‹๐‘ก).

To find the resultant wave, we add the two wave equations together: ๐ธ = ๐ธ1 + ๐ธ2.

๐ธ = 3๐‘๐‘œ๐‘ (๐›ผ1 โˆ’ 2๐œ‹๐‘ก) + 4๐‘๐‘œ๐‘ (๐›ผ2 โˆ’ 2๐œ‹๐‘ก).

Simplifying the equation further may require using trigonometric identities and properties.

This problem has been solved

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