Question

A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is a while at another instant velocity is 'v' and acceleration is 'β (0 < α < β). The distance between the two position is ______.

Options

• `("u"^2-"v"^2)/(alpha+beta)`

• `("u"^2+"v"^2)/(alpha+beta)`

• `("u"^2-"v"^2)/(alpha-beta)`

• `("u"^2+"v"^2)/(alpha-beta)`

Answer 1

A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is a while at another instant velocity is 'v' and acceleration is 'β (0 < α < β). The distance between the two position is `underlinebb(("u"^2-"v"^2)/(alpha+beta))`.

Explanation:

Let the distance be x when velocity is u and acceleration α.

And the distance y when velocity is v and acceleration β.

If ω is the angular frequency, then

α = ω²x and β = ω²y

∴ α + β = ω² (x + y)                      ...(i)

Also, u² = ω²A² - ω²x²

and v² = ω²A² - ω²y² 

⇒ v² - u² = ω²(x² - y²)

v² - u² = ω²(x - y)(x + y)             ...(ii)

From eqs. (i) and (ii),

v² - u² = (x - y) (α + β)

∴ x - y = `("v"^2-"u"^2)/(alpha+beta)` or y - x = `("u"^2-"v"^2)/(alpha+beta)`