If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to $2.0 \mat

Question

If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to $2.0 \mathrm{~m} / \mathrm{s}^{2}$ at any time, the angular frequency of the oscillator is equal to\

Professor PC Thomas Classes · Answer

(a) : Acceleration $=-\omega^{2}$ displacement\ $\omega^{2}=\frac{	ext { acceleration }}{	ext { displacement }}=\frac{2.0}{0.02} ; \omega^{2}=100$ or $\omega=10 \mathrm{rad} / \mathrm{s}$\