Question

A mass m at the end of a spring vibrates with a frequency of 0.92 Hz. When an additional 600 g mass is added to m, the frequency is 0.64 Hz. What is the value of m?

Answer 1

Start with the relation
f = [1/(2 pi)]sqrt(m/k)

The spring constant k remains constant

Since the frequency decreases by a factor of 1.4375, the mass must have increased by a factor 1.4375^2, which is 2.0664

Let M be the original mass that gave the 0.92 Hz frequency.

(M + 600)/600 = 2.0664
M/600 = 1.064
M = 638 g

You could also have solved this with two simultaneous equations and algebra, but this way is easier

Answer 2

that's not right, I think I'm gonna have to use 2 equations but i'm not sure how to do it.

Answer 3

I was careless and should have written
f = [1/(2 pi)] sqrt(k/m), but the following logic and final number I believe to be correct.

Answer 4

hmm Its not working out. this isnt the actual problem but one of the practice ones (online homework) but i keep trying this method with different numbers and they aren't working out. Do you know how to explain the way to do it with 2 equations?

Answer 5

0.92 = [1/(2 pi)] sqrt(k/m)
0.64 = [1/(2 pi)] sqrt[k/(m+0.6)],
where 0.6 is the added mass in kg.
Now divide the first equation by the second, to get rid of the unknown k

1.4375 = sqrt [(m + 0.6)/m]
2.0664 = (m + 0.6)/m = 1 + 0.6/m
1.0664 = 0.6/m
m = 0.563 kg

I made an algebraic mistake intitally, and should have written
(M + 600)/M = 2.0664
I was working with M in grams, which is OK when dealing with ratios, but I had the wrong denominator in the last equation.

So I not only wrote the original k/m ratio upside down, but also later made an algebraic error. I regret both errors and the confusion it caused you. I should have been more careful