Question

Two simple harmonic motions are represented by y1=4sin(4t2)y1=4sin(4t2) and y2=3cos(4t).Y2=3cos(4t). The resultant amplitude is :

Answer 1

Answer:

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Answer 2

The resultant amplitude of two given SHMs will be 5 units.

• Resultant amplitude is given by equation

        ((A1)∧2+(A2)∧2+ 2A1A2cos∆)^1/2     ...(1)

         where, ∆ = Phase difference

                     A1 & A2 = Amplitude of respective SHM.

• The general equation of SHM is y = Asin(ωt+Δ)

• y1 = 4sin(4t2) ⇒ A1 = 4
• y2 = 3cos(4t) ⇒ A2 = 3

Also,

• y2 = 3cos(4t) = 3sin(4t+(π/2)) ⇒ Δ = π/2

• Put values of A1, A2 & Δ in equation (1) we get,

       (4∧2+3∧2+ 2×4×3×cos(90))^1/2

        $\sqrt{16+9}$ = 5