A load of mass m falls from a height h on to the scale pan hung from a spring as shown in the adjoining figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is

A load of mass m falls from a height h on to the scale pan hung from a

1.	A load of mass m falls from a height h on to the scale pan hung from a spring as shown in the adjoining figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is
Answer» `(mg)/(k)` `(mg)/(k)SQRT(((1+2hk)/(mg)))` `(mg)/(k)+(mg)/(k)sqrt(((1+2hk)/(mg)))` `(mg)/(k)sqrt(((1+2hk)/(mg)-(mg)/(k)))` Solution :According to conservation principle `mgh=(1)/(2)kx_(0)^(2)-mgx_(0)` where, `x_(0)` is maximum elongation in spring (when PARTICLE is in its LOWEST EXTREME position). or `(1)/(2)kx_(0)^(2)-mgx_(0)-mgh=0` or `x_(0)^(2)-(2mg)/(k)x_(0)-(2mg)/(k)h=0` `therefore""x_(0)=((2mg)/(k)pmsqrt([((2mg)/(k))^(2)+4xx(2mg)/(k)h]))/(2)` `because` Amplitude=elongation in spring for lowest extreme position-elongation in spring for equilibrium position `=x_(0)-x_(1)=(mg)/(k)sqrt((1+(2hk)/(mg)))`