Help Period of oscillation of the mass

[andycl]

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!

 

[Dick]

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?

 

[ogb p]

Sure, I can provide some guidance and explanation for finding the period of oscillation in this scenario. First, let's define some terms and equations that will be helpful in solving this problem.

- Period (T): The time it takes for one complete oscillation or cycle.
- Frequency (f): The number of oscillations per unit time.
- Amplitude (A): The maximum displacement of the mass from its equilibrium position.
- Angular frequency (ω): The rate at which the mass oscillates, measured in radians per second.
- Equilibrium position (x = 0): The position where the mass is at rest with no external forces acting on it.

Now, to find the period of oscillation, we can use the equation T = 2π/ω, where ω = √(k/m) and k is the spring constant and m is the mass.

In this scenario, we are given the displacement (A = 0.16 m) and the position at t = 0.50 s (x = 0.08 m). We can use this information to find the angular frequency ω.

First, we need to find the position at t = 0, which is the equilibrium position. Since the mass is at rest at this position, we can set the equation for the net force on the mass to 0. This gives us the equation kx = 0, which means that x = 0 is the equilibrium position.

Now, we can use the given information to find the angular frequency. We know that A = 0.16 m and x = 0.08 m, so we can use the equation A = A₀cos(ωt) to find ω. Plugging in the values, we get:

0.16 = 0.08cos(ω*0.50)

Solving for ω, we get ω = 1.256 rad/s.

Now, we can plug this value into the equation T = 2π/ω to find the period of oscillation:

T = 2π/1.256 = 5.026 seconds.

Therefore, the period of oscillation of the mass is 5.026 seconds. I hope this explanation helps and clarifies the steps needed to solve this type of problem. Keep practicing and you'll get the hang of it!