Two equal masses M and N are suspended from the ends of two separate weightless springs having spring constants k_(1)andk_(2). If the maximum velocities of the two masses for their vertical oscillations are the same, what is the ratio of the amplitudes of vibration of M and N?

Two equal masses M and N are suspended from the ends of two separate w

1.	Two equal masses M and N are suspended from the ends of two separate weightless springs having spring constants k_(1)andk_(2). If the maximum velocities of the two masses for their vertical oscillations are the same, what is the ratio of the amplitudes of vibration of M and N?
Answer» SOLUTION :If angular frequency = `omega` and AMPLITUDE = A, then maximum velocity = `omegaA`. So for the TWO given masses, `omega_(1)A_(1)=omega_(2)A_(2)` or, `A_(1)/A_(2)=omega_(2)/omega_(1)=((2pi)/T_(2))/((2pi)/T_(1))=T_(1)/T_(2)=(2pisqrt(m_(1)/k_(1)))/(2pisqrt(m_(2)/k_(2)))` or, \`A_(1)/A_(2)=SQRT(k_(2)/k_(1))" "[becausem_(1)=m_(2)=m]`.