Question

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and the potential energies equal?

Answer 1

Using the Formula,

Kinetic Energy = 1/2 k (A² - x²)

Potential Energy = 1/2 kx²

Since, they are equal at a distance x.

∴ 1/2 kx² = 1/2 k (A² - x²)

x² = A² - x²

2x² = A²

x² = A²/2

x = ± A/√2

At this point, Kinetic and potential energy will be equal.

Hope it helps.

Answer 2

Sol.

r = 10cm Because,

K.E. = P.E.

So (1/2) m ω2 (r2– y2)

= (1/2) m ω2y2 r2 – y2 = y2

⇒ 2y2 = r2

⇒ y = r/√2

= 10/√2

= 5√2 cm form the mean position