Help Period of oscillation of the mass

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To determine the period of oscillation for a mass connected to a spring, the position can be expressed as x(t) = A*cos(omega*t), with A being the maximum displacement of 0.16 m. At t = 0.50 s, the mass is 0.08 m from equilibrium, allowing for the calculation of omega by substituting this value into the equation. Once omega is found, the period can be derived using the relationship between period and angular frequency (T = 2π/omega). This method provides a clear path to solving for the period of oscillation. Understanding these relationships is crucial for accurately calculating the dynamics of the system.
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A mass which is resting on a horizontal frictionless surface is connected to a fixed spring. The mass is displaced 0.16 m from its equilibrium position and released. At t = 0.50 s, the mass is 0.08 m from its equilibrium position (and has not passed through it yet).

What is the period of oscillation of the mass?

I know I am suppose to put what I have done so far but I have given up with the trying. Could somebody please give me some guidance and explain please.

Thank you!
 
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You can generally write the position of the mass wrt to equilibrium as x(t)=A*cos(omega*t+phi), where A is the maximum displacement and phi is the phase. Since you know you have maximum displacement at t=0 you can set phi=0 and A=0.16m. So x(t)=(0.16m)*cos(omega*t). Enough hints, now can you put the values at t=0.5 sec in and solve for omega? Knowing omega, can you figure out the period?
 
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