Question

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by y(t) = y₀ sin² ωt, where ‘y’ is measured from the lower end of the unstretched spring. Then ω is:

Options

• `1/2 sqrt(g/y_0)`

• `sqrt(g/y_0)`

• `sqrt(g/(2y_0))`

• `sqrt((2g)/(y_0))`

Answer 1

`bb(sqrt(g/(2y_0)))`

Explanation:

y = y₀ sin² ωt

⇒ y = `y_0/2 (1 - cos 2 ωt)  ...(∵ sin^2 ωt = (1 - cos 2 ωt)/2)`

⇒ `y - y_0/2 = (-y_0)/2` cos ωt

So from this equation, we can say the mean position is shifted by `y_0/2` distance and the frequency of this SHM is 2ω.

So, at equilibrium `(ky_0)/2` = mg

⇒ `k/m = (2g)/y_0`

Also, spring constant k = m(2ω)²

2ω = `sqrt(k/m)`

2ω = `sqrt((2g)/y_0)`

ω = `1/2 sqrt((2g)/y_0)`

ω = `sqrt(g/(2y_0))`