When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4Hz. Find the ratio m2/m1 of the masses.
To find the ratio of the masses, we can use the relationship between frequency and mass in simple harmonic motion.
The formula for the frequency of a mass-spring system is given by:
f = 1 / (2π) * √(k / m),
where f is the frequency, k is the spring constant, and m is the mass.
Let's assume that m1 is the initially hung mass and m2 is the additional mass. With only m1, the frequency is 12Hz, so we can write the equation as:
12 = 1 / (2π) * √(k / m1).
Now, when m2 is added, the frequency becomes 4Hz:
4 = 1 / (2π) * √(k / (m1 + m2)).
We need to find the ratio m2/m1. To simplify the calculation, we can divide the second equation by the first equation:
(4 / 12) = (√(k / (m1 + m2))) / (√(k / m1)).
Simplifying further:
1/3 = √(m1 / (m1 + m2)).
Now, we can square both sides of the equation to eliminate the square root:
(1/3)^2 = m1 / (m1 + m2).
1/9 = m1 / (m1 + m2).
To isolate m2/m1, we can rearrange the equation as follows:
(1/9) = m1 / (m1 + m2).
Cross-multiplying gives:
m2 + m1 = 9m1.
m2 = 9m1 - m1.
m2 = 8m1.
Therefore, the ratio of m2/m1 is 8/1 or simply 8.