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

पर्याय

• k₁/k₂

• \[\sqrt{k_1 / k_2}\]

• k₂/k₁

• _{\[\sqrt{k_2 / k_1}\]}

Answer 1

\[\sqrt{\frac{k_2}{k_1}}\]

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,

\[i . e . v_A = v_B \]

\[\text { or }, \]

\[ A \omega_A = B \omega_B \]

\[ \Rightarrow A\sqrt{\frac{k_1}{m_A}} = B\sqrt{\frac{k_2}{m_B}}\]

\[ \Rightarrow A\sqrt{\frac{k_1}{m}} = B\sqrt{\frac{k_2}{m}} ( m_A = m_B = m)\]

\[ \Rightarrow \frac{A}{B} = \sqrt{\frac{k_2}{k_1}}\]