Two masses m-1 and m-2 are suspended together by a mass spring of constant K-When the masses are in equilibrium- m-1 is removed without disturbing the system-Then the angular frequency of oscillation of m-2 is-sqrt -dfrac -K-m-1-sqrt -dfrac -K-m-2-sqrt -dfrac -K-m-1-m-2-sqrt -dfrac -K-m-1-m-2-

Question

Two masses  m-1 and  m-2 are suspended together by a mass spring of constant K-When the masses are in equilibrium-  m-1 is removed without disturbing the system-Then the angular frequency of oscillation of  m-2 is-sqrt -dfrac -K-m-1-sqrt -dfrac -K-m-2-sqrt -dfrac -K-m-1-m-2-sqrt -dfrac -K-m-1-m-2-

Toppr Admin · Accepted Answer

After removal of m1 oscillations are only due to m2- So the angular frequency of oscillations of m2 is-x3C9-x221A-Km2

Two masses m1 and m2 are suspended together by a mass less spring of constant K.When the masses are in equilibrium, m1 is removed without disturbing the system.Then the angular frequency of oscillation of m2 is:√Km1√Km2√K(m1+m2)√K(m1−m2)

After removal of m1 oscillations are only due to m2. So the angular frequency of oscillations of m2 isω=√Km2

After removal of $m_{1}$ oscillations are only due to $m_{2}$ . So the angular frequency of oscillations of $m_{2}$ is
$ω = \sqrt{\frac{K}{m_{2}}}$