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NEET PHYSICSMedium

When two displacements represented by y1=asin(ωt)y_1 = a \sin(\omega t) and y2=bcos(ωt)y_2 = b \cos(\omega t) are superimposed, the motion is:

A

not a simple harmonic

B

simple harmonic with amplitude a/b

C

simple harmonic with amplitude a2+b2\sqrt{a^2+b^2}

D

simple harmonic with amplitude (a+b)/2

Step-by-Step Solution

The superposition of two simple harmonic motions (SHM) with the same angular frequency ω\omega results in a new simple harmonic motion. The resultant displacement yy is the sum of the individual displacements: y=y1+y2=asin(ωt)+bcos(ωt)y = y_1 + y_2 = a \sin(\omega t) + b \cos(\omega t) Using the trigonometric identity cos(ωt)=sin(ωt+π2)\cos(\omega t) = \sin(\omega t + \frac{\pi}{2}), we see that the two waves have a phase difference of ϕ=π2\phi = \frac{\pi}{2}. The resultant amplitude RR for the superposition of two waves with amplitudes aa and bb and phase difference ϕ\phi is given by: R=a2+b2+2abcosϕR = \sqrt{a^2 + b^2 + 2ab \cos\phi} Substituting ϕ=π2\phi = \frac{\pi}{2} (since cosπ2=0\cos \frac{\pi}{2} = 0): R=a2+b2R = \sqrt{a^2 + b^2} Thus, the resulting motion is simple harmonic with an amplitude of a2+b2\sqrt{a^2 + b^2}. This concept is consistent with the principles of phasor addition used in analyzing AC circuits and oscillations .

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