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JEE PHYSICS MCQ: OSCILLATIONS

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Analyzing +4 | -1
The displacement of an oscillator is given by 
x=asin⁡ωt+bcos⁡ωt\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \cos \omega \mathrm{t}
where a,b\mathrm{a}, \mathrm{b} and ω\omega, are constant. Then -
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Correct option: D
motion is neither simple harmonic nor periodic
Option C (motion is simple harmonic as well as periodic) is correct because the given equation represents the superposition of two SHM's with the same frequency, resulting in SHM.

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