Calculating Amplitude and Phase for Superimposed Harmonic Oscillators

[kidia]

Can anybody give me the hint where to start on this question?

Two simple harmonic oscillators of the same frequency and in the same direction having amplitudes 5 mm and 3 mm, respectively and the phase of the second component relative to the first is 30°, are superimposed. Find the amplitude of the resultant oscillation and its phase relative to the first component.

 

[Chi Meson]

Whether or not it will suffice as "properly done," try manually graphing these waves on graph paper (two cycles of each wave, max), and adding their amplitudes to find the superimposed amplitude. It will give you insight to what is going on.

If you need to show the math, begin by finding the expression for amplitude for each wave as a funtion of wt.

 

[kyle8921]

To calculate the amplitude and phase for superimposed harmonic oscillators, we first need to understand the concept of superposition. Superposition is the principle that states that when two or more waves or oscillations are combined, the resulting wave is the sum of the individual waves.

In this case, we have two simple harmonic oscillators with the same frequency and direction but different amplitudes and a phase difference of 30°. To find the amplitude of the resultant oscillation, we can use the formula A = √(A1^2 + A2^2 + 2A1A2cosϕ), where A1 and A2 are the amplitudes of the individual oscillators and ϕ is the phase difference between them.

Substituting the given values, we get A = √(5^2 + 3^2 + 2(5)(3)cos30°) = √(25 + 9 + 30) = √64 = 8 mm.

To find the phase of the resultant oscillation relative to the first component, we can use the formula tanϕ = (A2sinϕ)/(A1 + A2cosϕ). Substituting the given values, we get tanϕ = (3sin30°)/(5 + 3cos30°) = 0.5/6.5 = 0.0769. Taking the inverse tangent, we get ϕ = 4.4°.

Therefore, the amplitude of the resultant oscillation is 8 mm and its phase relative to the first component is 4.4°. This means that the resulting oscillation will have an amplitude of 8 mm and will be shifted by 4.4° in the direction of the second component.