First slide

Simple hormonic motion

Question

The displacement of two identical particles executing SHM are represented by equations $x_{1} = 4 \sin (10 \sqrt{2} t + (π / 6)) a n d x_{2} = 5 \sqrt{2} \cos ω t$ . For what value of $ω$ energy of both the particles is same?

Easy

Solution

$\begin{array}{l} \frac{1}{2} m {ω_{1}}^{2} {A_{1}}^{2} = \frac{1}{2} m {ω_{2}}^{2} {A_{2}}^{2} \\ 200 \times 4^{2} = ω^{2} 50 \\ ∴ ω = 8 u n i t s \end{array}$

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