To solve for the ratio m2/m1 of the masses, we need to use the formula for the frequency of vertical simple harmonic motion of a mass-spring system:
f = 1 / (2π) * √(k/m)
where f is the frequency, k is the spring constant, and m is the mass.
Given that the frequency with just m1 is 12 Hz, we can rewrite the formula as:
12 = 1 / (2π) * √(k / m1)
Squaring both sides of the equation, we get:
144 = (1 / (2Ï€))^2 * (k / m1)
Similarly, for the system with both m1 and m2, the frequency is 4 Hz:
4 = 1 / (2π) * √(k / (m1 + m2))
Again, squaring both sides:
16 = (1 / (2Ï€))^2 * (k / (m1 + m2))
Now, we can compare the two equations:
144 = (1 / (2Ï€))^2 * (k / m1)
16 = (1 / (2Ï€))^2 * (k / (m1 + m2))
Dividing the second equation by the first equation:
16 / 144 = (k / (m1 + m2)) / (k / m1)
1 / 9 = (k / (m1 + m2)) * (m1 / k)
Canceling out the k's:
1 / 9 = m1 / (m1 + m2)
Cross multiplying:
m1 + m2 = 9m1
Now, rearranging the equation:
m2 = 8m1
Therefore, the ratio m2/m1 is 8:1.