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) $\sqrt{k_1/k_2}$

(c) k₂/k₁

(d) $\sqrt{k_2/k_1}$

Answer 1

(d)$\sqrt{\frac{k_2}{k_1}}$

Maximum velocity, v = Aω
where A is amplitude and ω is the angular frequency.

Further, ω = $\sqrt{\frac{k}{m}}$
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,

$$\begin{gathered} i.e.v_A=v_B \\ \mathrm{or}, \\ A{\omega }_A=B{\omega }_B \\ \Rightarrow A\sqrt{\frac{k_1}{m_A}}=B\sqrt{\frac{k_2}{m_B}} \\ \Rightarrow A\sqrt{\frac{k_1}{m}}=B\sqrt{\frac{k_2}{m}}(m_A=m_B=m) \\ \Rightarrow \frac{A}{B}=\sqrt{\frac{k_2}{k_1}} \end{gathered}$$