Find the resultant of the superposition of two harmonic waves in the form ๐ธ = ๐ธ0 ๐๐๐ ๐๐๐ (๐ผ โ ๐๐ก) ,with amplitudes of 3 and 4 units and phases of ๐6 and ๐2 respectively. Both waves have a period of 1 s
Question
Find the resultant of the superposition of two harmonic waves in the form ๐ธ = ๐ธ0 ๐๐๐ ๐๐๐ (๐ผ โ ๐๐ก) ,with amplitudes of 3 and 4 units and phases of ๐6 and ๐2 respectively. Both waves have a period of 1 s
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 ๐ธ1 = ๐ธ0๐๐๐ (๐ผ1 โ ๐๐ก) and ๐ธ2 = ๐ธ0๐๐๐ (๐ผ2 โ ๐๐ก), where ๐ธ0 is the amplitude, ๐ผ is the phase, and ๐ is the angular frequency.
For the first wave, ๐ธ1, the amplitude is 3 units and the phase is ๐/6. The angular frequency, ๐, can be calculated using the formula ๐ = 2๐/T, where T is the period. In this case, the period is 1 second, so ๐ = 2๐/1 = 2๐ rad/s.
Substituting the values into the wave equation, ๐ธ1 = 3๐๐๐ (๐ผ1 โ 2๐๐ก).
For the second wave, ๐ธ2, the amplitude is 4 units and the phase is ๐/2. Using the same formula for ๐, we find ๐ = 2๐ rad/s.
Substituting the values into the wave equation, ๐ธ2 = 4๐๐๐ (๐ผ2 โ 2๐๐ก).
To find the resultant wave, we add the two wave equations together: ๐ธ = ๐ธ1 + ๐ธ2.
๐ธ = 3๐๐๐ (๐ผ1 โ 2๐๐ก) + 4๐๐๐ (๐ผ2 โ 2๐๐ก).
Simplifying the equation further may require using trigonometric identities and properties.
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