Question

Two Bodies M And N Of Equal Masses Are Suspended From Two Separate Massless Springs Of Force Constants K1 And K2 Respectively If The Two Bodies Oscillate Vertically Such That Their Maximum Velocities Are Equal , The Ratio Of The Amplitude M To That Of N Is_____.

Answer 1

Let $\sf{A_1}$ and $\sf{A_2}$ be amplitudes of M and N respectively.

Similarly, let $\sf{\omega_1}$ and $\sf{\omega_2}$ be angular speeds of M and N respectively.

Maximum velocity is given by,

$\sf{\longrightarrow v_{max}=A\omega}$

Given that maximum velocities of M and N each are equal.

$\sf{\longrightarrow A_1\omega_1=A_2\omega_2}$

$\sf{\longrightarrow\dfrac{A_1}{A_2}=\dfrac{\omega_2}{\omega_1}\quad\quad\dots(1)}$

But, angular speed is given by,

$\sf{\longrightarrow \omega=\sqrt{\dfrac{K}{M}}}$

Since M and N have equal masses.

$\sf{\longrightarrow \omega\propto\sqrt K}$

Therefore,

$\sf{\longrightarrow \dfrac{\omega_1}{\omega_2}=\sqrt{\dfrac{K_2}{K_1}}}$

Hence (1) becomes,

$\sf{\longrightarrow\underline{\underline{\dfrac{A_1}{A_2}=\sqrt{\dfrac{K_2}{K_1}}}}}$