Question 339: Oscillations

Question

A body of mass '$$m$$' performs linear S.H.M. given by equation $$x=P \sin \omega t+Q \sin \left(\omega t+\frac{\pi}{2}\right)$$. The total energy of the particle at any instant is

Accepted Answer

To find the total energy of a particle performing Simple Harmonic Motion (S.H.M.), we start with the given equation of motion:
$$x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}ight)$$
First, we simplify the given equation using trigonometric identities. The identity for $$\sin \left(\omega t + \frac{\pi}{2}ight)$$ can be rewritten as $$\cos \omega t$$ because $$\sin \left(	heta + \frac{\pi}{2}ight) = \cos 	heta$$. Hence, the motion equation becomes:
$$x = P \sin \omega t + Q \cos \omega t$$
To find the total energy, we need to consider both kinetic and potential energies of the system. The total energy in S.H.M. is given by the sum of potential and kinetic energies, which remains constant and can be represented as:
$$E = \frac{1}{2} m \omega^2 A^2$$
where $$A$$ is the amplitude of the motion. In this case, the total amplitude of the motion given by the resultant of $$P \sin \omega t$$ and $$Q \cos \omega t$$ is actually the resultant vector magnitude of $$P$$ and $$Q$$, since they can be seen as perpendicular components (similar to a Pythagorean theorem scenario), and hence the resultant or effective amplitude $$A$$ is:
$$A = \sqrt{P^2 + Q^2}$$
Thus, substituting $$A = \sqrt{P^2 + Q^2}$$ into the equation for total energy, we get:
$$E = \frac{1}{2} m \omega^2 (\sqrt{P^2 + Q^2})^2$$
$$E = \frac{1}{2} m \omega^2 (P^2 + Q^2)$$
Therefore, the correct option is:
Option C: $$\frac{1}{2} m \omega^2\left(P^2+Q^2ight)$$

Answer

$$\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{PQ}$$

Answer

$$\frac{1}{2} \frac{\mathrm{m} \omega^2}{\mathrm{P}^2 \mathrm{Q}^2}$$

Answer

$$\frac{1}{2} m \omega^2\left(P^2+Q^2\right)$$

Answer

$$\frac{1}{2} m \omega^2 p^2 Q^2$$

Oscillations

A body of mass '$m' performs linear S.H.M. given by equation x=P \sin \omega t+Q \sin \left(\omega t+\frac{\pi}{2}\right)$. The total energy of the particle at any instant is

Choose the correct answer: