Simple Harmonic Motion problem with spring mass system

[physicsuwa]

Homework Statement

A 50g mass is attached to a spring and undergoes simple harmonic motion. It's maximum acceleration is 15m/s and maximum speed is 3.5m/s. Determine a)angular frequency, b) spring constant, c) amplitude.

Homework Equations

ω = √k/m
X(t) = A*cos(ωt + ∅)

The Attempt at a Solution

Tried but I seem to be getting nowhere. Would really appreciate if someone pointed me in the right direction.

 

[Dick]

Homework Statement

A 50g mass is attached to a spring and undergoes simple harmonic motion. It's maximum acceleration is 15m/s and maximum speed is 3.5m/s. Determine a)angular frequency, b) spring constant, c) amplitude.

Homework Equations

ω = √k/m
X(t) = A*cos(ωt + ∅)

The Attempt at a Solution

Tried but I seem to be getting nowhere. Would really appreciate if someone pointed me in the right direction.

You need to show what you tried. Try taking derivatives of X(t) to find expressions for the velocity and acceleration.

 

[physicsuwa]

You need to show what you tried. Try taking derivatives of X(t) to find expressions for the velocity and acceleration.

I did that actually. I got expressions for both but I am missing amplitude so can't gain much from that. I tried to used f=-kx and substitute that into the ω = √k/m but that didn't work out either as I am missing x. Anyway. I just need a direction or that clue that will just crack this open for me. I appreciate the help.

 

[Dick]

I did that actually. I got expressions for both but I am missing amplitude so can't gain much from that. I tried to used f=-kx and substitute that into the ω = √k/m but that didn't work out either as I am missing x. Anyway. I just need a direction or that clue that will just crack this open for me. I appreciate the help.

Just show what you got. It will turn out that you can cancel the amplitude with the correct equations for the maximum velocity and acceleration and find ω. Divide the maximum acceleration by the maximum velocity.

 

[physicsuwa]

Ok. Well what i have done is in regards to the velocity and acceleration is differentiated the distance to get velocity and differentiated that to get acceleration.
v = ω*A*sin(ωt+∅) and for acceleration is a=-ω²*A*cos(ωt+∅). I am quite sure that's right but let me know if its not.

and for angular frequency which is ω=√k/m which i mentioned before and Period or T = 2π*√m/k

 

[BvU]

Differentiating the displacement x(t) is right. I would expect a minus sign for v too.

So now you have expressions and values for maximum v and maximum a. Two equations with two unknowns (A and omega). Bingo.

Oh, and: Welcome to PF, Uwa !

 

[Dick]

Ok. Well what i have done is in regards to the velocity and acceleration is differentiated the distance to get velocity and differentiated that to get acceleration.
v = ω*A*sin(ωt+∅) and for acceleration is a=-ω²*A*cos(ωt+∅). I am quite sure that's right but let me know if its not.

and for angular frequency which is ω=√k/m which i mentioned before and Period or T = 2π*√m/k

What are the maximum values of v and a?

 

[physicsuwa]

What are the maximum values of v and a?

I gave them in the original question. but perfectly happy to repeat. Max Speed = 3.5m/s and max acceleration is 15m/s^2

 

[Dick]

I gave them in the original question. but perfectly happy to repeat. Max Speed = 3.5m/s and max acceleration is 15m/s^2

I meant what are they in terms of $\omega$ and $A$!

 

[physicsuwa]

Sorry. I don't understand what you mean by "in terms" of ω and A

 

[physicsuwa]

Dick. I am sorry man. I just saw it and it was literally a Eureka moment on the train. I don't know how I could have been so stupid. Thanks for sticking with me. Appreciate all the help.

 

[Dick]

Dick. I am sorry man. I just saw it and it was literally a Eureka moment on the train. I don't know how I could have been so stupid. Thanks for sticking with me. Appreciate all the help.

You're welcome. Always worthwhile when somebody gets it.