Question

Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k₁/k₂
(b) √(k₁/k₂)
(c) k₂/k₁
(d) √(k₂/k₁)

Answer 1

Using the Formula,

v² = ω²(A² - x²), where A is the amplitude, w is the angular frequency and x is the distance from the origin.

Velocity is maximum, when x = 0,

therefore, v² = ω²A²

∴ v = ±ωA

ω = k/m

For body A,

v₁ = ω₁A₁

v₁ = A₁√k₁/m

For body B,

v₂ = A₂√k₂/m

Now, v₁ = v₂

therefore,

A₁√k₁/m = A₂√k₂/m

∴ A₁/A₂ = √(k₂/k₁)

Hence, option (d). is correct.

Hope it helps .

Answer 2

Let the amplitudes A and B be A1 and A2 respectively.

Maximum velocity for a body performing SHM is given as:

v(max) = ωA

where, ω = √(k/m)

So, Maximum velocity of A = max. velocity of B

$or, \omega_AA_1 = \omega_BA_2$

$\sqrt{\dfrac{k_1}{m}}.A_1 = \sqrt{\dfrac{k_2}{m}}.A_2\\\\\boxed{\dfrac{A_1}{A_2} = \sqrt{\dfrac{k_2}{k_1}}}$

Hence, Option d) √(k₂/k₁) is correct.