Amplitude of oscillation for SHM

[Saptarshi Sarkar]

Homework Statement

A spring is attached to the roof of an elevator and a mass m is attached to the lower end of the spring. When the elevator is not moving, the mass undergoes a vertical SHM with frequency 2.5rad/s. Now, when the elevator is moving downwards with velocity 0.5m/s, the mass is not undergoing any SHM. Now, if the elevator suddenly stops, what will be the amplitude of oscillation of the mass m?

Relevant Equations

y = ACosωt + BSinωt

From the first part of the question, I was able to get the value of ω which will be the same for the next SHM.

But, I am having difficulties solving for the amplitude as I can't find the boundary conditions required to get the amplitude.

 

[Cutter Ketch]

Just before the elevator stops the mass is not moving in the elevator frame. When the elevator stops suddenly how fast is the mass moving relative to the elevator? And, considering it wasn’t oscillating a moment before, what location in the oscillation does it start?

 

[Saptarshi Sarkar]

Just before the elevator stops the mass is not moving in the elevator frame. When the elevator stops suddenly how fast is the mass moving relative to the elevator? And, considering it wasn’t oscillating a moment before, what location in the oscillation does it start?

When the elevator stops, the mass should move with velocity 0.5m/s wrt the elevator frame. The mass starts moving from the centre of SHM (y=0)

So, that means

y(0)=0
y'(0)=0.5

Is this right?

 

[Cutter Ketch]

Yep, where you have defined down as the positive y direction (perfectly acceptable so long as you are consistent through the rest of the calculation)

 

[Cutter Ketch]

Also, y(0) = 0 should allow you to simplify your equation.