Question

A body of mass m performs linear S.H.M. given by equation, x = P sin cot + Q sin${\displaystyle (\omega \text{t}+\frac{\pi }{2})}$. The total energy of the particle at any instant is ______.

Options

• ${\displaystyle \frac{1}{2}\text{m}\omega ({\text{P}}^2+{\text{Q}}^2)}$

• ${\displaystyle \frac{1}{2}\text{m}{\omega }^2/\text{PQ}}$

• ${\displaystyle \frac{1}{2}\text{m}{\omega }^2\text{PQ}}$

• ${\displaystyle \frac{1}{2}\text{m}{\omega }^2{\text{P}}^2{\text{Q}}^2}$

Answer 1

A body of mass m performs linear S.H.M. given by equation, x = P sin cot + Q sin${\displaystyle (\omega \text{t}+\frac{\pi }{2})}$. The total energy of the particle at any instant is ${\displaystyle \underline{\frac{1}{2}\text{m}\omega ({\text{P}}^2+{\text{Q}}^2)}}$.

Explanation:

x = P sin cot + Q sin${\displaystyle (\omega \text{t}+\frac{\pi }{2})}$

= P sin ωt + Q cos ωt

Let P = A cos θ    Q = A sin θ

∴ P² + Q² = A²

∴ x = A sin (ωt + θ)

${\displaystyle {\text{E}}_{\text{total}}=\frac{1}{2}\text{m}{\omega }^2{\text{A}}^2}$

${\displaystyle =\frac{1}{2}\text{m}{\omega }^2({\text{P}}^2+{\text{Q}}^2)}$