Question

A mass m₁ connected to a horizontal spring performs SHM with amplitude A. While mass m₁ is passing through mean position, another mass m₂ is placed on it so that both the masses move together with amplitude A₁. The ratio of `"A"_1/"A"` is ______. (m₂ < m₁)

विकल्प

• `["m"_1/("m"_1+"m"_2)]^(1/2)`

• `[("m"_1+"m"_2)/"m"_1]^(1/2)`

• `["m"_2/("m"_1+"m"_2)]^(1/2)`

• `[("m"_1+"m"_2)/"m"_2]^(1/2)`

Answer 1

A mass m₁ connected to a horizontal spring performs SHM with amplitude A. While mass m₁ is passing through mean position, another mass m₂ is placed on it so that both the masses move together with amplitude A₁. The ratio of `"A"_1/"A"` is `underlinebb(["m"_1/("m"_1+"m"_2)]^(1/2))`. (m₂ < m₁)

Explanation:

For a given oscillating mass, potential energy is given by

PE = `1/2`kx²

For a body oscillating at x = A, maximum energy is given by

E_max = `1/2`kA²

Also at mean position x = 0

So, E = 0

∴ A ∝ `1/sqrt"m"`            ...(i)

When another mass m₂ is placed on mass m₁. Then, total mass becomes (m₁ + m₂) and at this point E = 0 as x = 0.

When they reach at x = A,

A₁ ∝ `1/sqrt("m"_1+"m"_2)`         ...(ii)

Dividing eq. (ii) by eq. (i),

`"A"_1/"A"=["m"_1/("m"_1+"m"_2)]^(1/2)`