Question

Two simple harmonic motions, \( y _{1}= a \sin \omega t \) and \( y _{2}=2 a \sin \left(\omega t +\frac{2 \pi}{3}\right) \) are superimposed on a particle of mass \( m \). The maximum kinetic energy of the particle is: 1. \( \frac{1}{2} m \omega^{2} a^{2} \) 2. \( \frac{5 m \omega^{2} a ^{2}}{4} \) 3. \( \frac{3}{2} m \omega^{2} a^{2} \) 4. Zero

Answer 1

Correct option is (a)  \(\frac12\)mω²a²

y₁ = a sin ωt

y₂ = 2a sin(ωt + 2π/3)

then

Resultant amplitude = \(\sqrt{a^2+(2a)^2+2\times2a\times a cos 120^\circ}\)

 = \(\sqrt{a^2+4a^2-4a^2} \) 

 = a

Maximum kinetic energy E = \(\frac12\)mω²a²