Question

Two bodies A and B of equal mass are suspended from two separated massless spring of spring constant k1 and k2 respectively if the bodies oscillate such that their maximum velocities are equal the ratio of the amplitude of A to that B is




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Answer 1

Answer:

I don’t know the answer because I am not a nerdish .

Explanation:

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.