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Asked by Shreyansh Last Modified
Answer As a seasoned tutor on UrbanPro, I'd be delighted to guide you through this physics problem.
When a mass attached to a spring is free to oscillate without friction or damping, we can model its motion using simple harmonic motion (SHM) principles. Let's break down the problem step by step.
At time t=0t=0, the mass is pulled to a distance x0x0 from its equilibrium position and given an initial velocity v0v0 towards the center.
In SHM, the equation governing the motion of the mass is: x(t)=Acos(ωt+ϕ)x(t)=Acos(ωt+ϕ)
Where:
Since the mass is initially displaced from equilibrium and given an initial velocity, we'll need to determine the amplitude AA in terms of ωω, x0x0, and v0v0.
The general equation for the velocity of an object undergoing SHM is: v(t)=−Aωsin(ωt+ϕ)v(t)=−Aωsin(ωt+ϕ)
At t=0t=0, the velocity of the mass is v0v0 towards the center, so: v(0)=−Aωsin(ϕ)=v0v(0)=−Aωsin(ϕ)=v0
At t=0t=0, the displacement of the mass is x0x0 from equilibrium, so: x(0)=Acos(ϕ)=x0x(0)=Acos(ϕ)=x0
We now have two equations: Acos(ϕ)=x0Acos(ϕ)=x0 −Aωsin(ϕ)=v0−Aωsin(ϕ)=v0
We can solve these equations simultaneously to find AA and ϕϕ: tan(ϕ)=−v0ωx0tan(ϕ)=−ωx0v0
Once we find ϕϕ, we can substitute it back into one of the equations to find AA: A=x0cos(ϕ)=x01+(v0ωx0)2A=cos(ϕ)x0=1+(ωx0v0)2
x0
Thus, we've determined the amplitude AA in terms of the parameters ωω, x0x0, and v0v0.
In conclusion, using the principles of SHM and the given initial conditions, we've found the amplitude of the resulting oscillations in terms of ωω, x0x0, and v0v0. If you need further clarification or assistance, feel free to ask!
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