Download the app

Questions  

 A body of mass M is released from a height h to a scale pan hung from a spring as shown in fig (9). The spring 

constant of the spring is k, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is

a
mgk1−2hkmg
b
mgk
c
mgk1+1+2hkmg
d
mgk−mgk1−2hkmg

detailed solution

Correct option is C

Let y be the displacement of the pan when the mass m hits it. Decrease in potential energy of mass =mg(h+y),y= amplitude. Increase in elastic potential energy of spring=12ky2 ∴ mg(h+y)=12ky2 or y2−2mgry−2mghk=0 y2−2mgry−2mghk=0 y=(2mg/k)±(2mg/k)2+4(2mgh/k)2 Reiecting negative sign, the total extension is y=mgk+mgk1+2hkmg =mgk1+1+2hkmg

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

The identical springs of constant ‘K’ are connected in series and parallel as shown in figure A mass M is suspended from them. The ratio of their frequencies of vertical oscillations will be


phone icon
whats app icon