Question

The displacement of two identical particles executing S.H.M are represented by equations
$$x_1=4\sin \left( 10t+\dfrac{\pi}{6}\right)$$ and $$x_2=5\cos \omega t$$
For what value of $$\omega$$ energy of both the particles is same?

Answer 1

Correct option is D. $$8$$ unit
Energy by a particle in SHM is given by

$$E=\dfrac{1}{2}m\omega^2A^2$$; where A is amplitude.

Given amplitude $$A_1=4$$ and $$A_2=5$$

$$\omega_1=10$$ and $$\omega_2=\omega$$

$$E_1=\dfrac{1}{2}m_1\omega_1^2A_1^2$$

$$E_2=\dfrac{1}{2}m_2\omega_2^2A_2^2$$

according to question

$$E_1=E_2$$

$$\dfrac{1}{2}m_1\omega_1^2A_1^2$$$$=\dfrac{1}{2}m_2\omega_2^2A_2^2$$

putting values

$$\dfrac{1}{2}m10^2\times4^2$$$$=\dfrac{1}{2}m\omega_2^25^2$$

$$1600=25\omega^2$$

$$\omega=8$$ unit