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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

The displacement of two particles of same mass executing simple harmonic motions are

x_1=16 \sin \left(10 t+\frac{\pi}{2}\right)

and

x_2=25 \cos (\omega t)

respectively.
The value of '

\omega

' for which the energies of both the particles remain the same is

View answer and explanation

Correct option: D

6.4 unit

Option D (6.4 unit) is correct because it is the value of

\omega

that equates the energies of the two particles, derived by setting

E_1 = E_2

and solving for

\omega_2

.

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