when an object of mass M1 is hung on a vertical spring and set vertical simple harmonic motion, its frequency is 12 Hz. 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.
Dont you have an equation of the period on a spring? Remember that period is 1/frequency.
I will be happy to check your thinking.
w^2 = k/m
w = 2 pi f
=6.285 X 12hz = 5688
m1 = 5688/k
following the same steps
m1 + m2 = 632/k
i am stuck...
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 / (2π))^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 / (2π))^2 * (k / (m1 + m2))
Now, we can compare the two equations:
144 = (1 / (2π))^2 * (k / m1)
16 = (1 / (2π))^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.