How Do You Calculate Mass and Spring Constant from Oscillation Period Changes?

AI Thread Summary
To determine the unknown mass M and spring constant K from the oscillation period changes, the initial period of 1 second and the increased period of 1.6 seconds after adding 200 grams are critical. The formula T=2π√(M/K) is used to relate mass and spring constant to the oscillation period. The discussion focuses on solving the equation 1/1.6 = √(M/(M+0.2)), emphasizing that mass cannot cancel out when squared. Participants guide each other through algebraic manipulation to isolate M, ultimately leading to a linear equation for M. The conversation highlights the importance of careful algebraic steps in solving for two variables.
izmeh
In a lab write up, i am given the following problem.

An unkown mass M is found to have a period of oscillation of 1 second using a spring of unkown spring constant K. When 200gms were added to the initial mass, the period of oscillation increased by 0.6 seconds. Determin the unkwon mass M(gms) and the Spring Constant K(N/m).

I am not very good at working with problems that have 2 variables. I have come up with a formula however.

T=2pi(sq.rt[M/K])
1 = 2pi(sq.rt[M/K])
1.6 = 2pi(sq.rt[m+.2]/K]
1/1.6 = sq.rt[m/m+.2]

however, i can not for the life of me figure out how to get the answer for M for this.

Is anyone willing to help me?
 
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square both sides & solve for m
 
m cancels out
 
Try it again. It doesn't cancel out.

You're now starting from this line right?
1/1.6 = sq.rt[m/m+.2]
How can m cancel out? When you square it, the right side becomes m/(m+.2).

(If you still can't get it, post what you're doing so we can find the error.)
 
.625² = sq.rt[m/m+.2]
.391 = m/m+0.2
.391(m+.2) = M
.391m + .078 = m

and so on...
 
Yes, so?

Now you have a simple linear equation for m. There's no way that m cancels out. Finish it & you'll have your solution.
 
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