Two particles of same mass are in simple harmonic motion, equation of their displacement x 1 = 4 sin ( 10 t + 6 π ) and x 2 = 5 cos ω t . Find the value of ω (in rad/s) for which total energy of both particles is same.

Equal energies: 2 1 m A 1 2 ω 1 2 = 2 1 m A 2 2 ω 2 2 . 16 × 100 = 25 ω 2 2 gives ω 2 = 8 rad/s.

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Two particles of same mass are in simple harmonic motion, equation of their displacement $x_{1} = 4 sin (10 t + \frac{π}{6})$ and $x_{2} = 5 cos ω t$ . Find the value of $ω$ (in rad/s) for which total energy of both particles is same.

Answer:

Solution:

Equal energies: $\frac{1}{2} m A_{1}^{2} ω_{1}^{2} = \frac{1}{2} m A_{2}^{2} ω_{2}^{2}$ .

$16 \times 100 = 25 ω_{2}^{2}$ gives $ω_{2} = 8$ rad/s.

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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