When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 14.2 Hz. When another object of mass m2 is hung on the spring along with the first object, the frequency of the motion is 3.90 Hz. Find the ratio m2/m1 of the masses.
To solve this problem, we need to use the concept of simple harmonic motion and the equation for the frequency of a mass-spring system.
Let's denote the ratio of masses as m2/m1 = x, where x is an unknown value we want to find.
The equation for the frequency of an object undergoing simple harmonic motion with a spring is:
f = 1 / (2π) * √(k / m)
Where:
f is the frequency of the motion,
k is the spring constant, and
m is the mass of the object.
Let's start by finding the spring constant, k. We'll use the second scenario where both objects are hung on the spring.
Let's assume the frequency of the motion with only the second object (mass m2) is f2 = 3.90 Hz.
Using the equation for the frequency, we have:
f2 = 1 / (2π) * √(k / m2)
Rearranging the equation, we can solve for k:
k = (2π * f2)² * m2
Next, let's find the spring constant, k, for the first scenario where only the first object (mass m1) is hung on the spring.
Using the equation for the frequency, we have:
f1 = 1 / (2π) * √(k / m1)
Rearranging the equation, we can solve for k:
k = (2π * f1)² * m1
The frequency in this scenario is given as f1 = 14.2 Hz.
Now that we have expressions for k in terms of m1 and m2, we can equate them and solve for the ratio of masses, x:
(2π * f1)² * m1 = (2π * f2)² * m2
Simplifying the equation:
(m2 / m1) = ((2π * f1) / (2π * f2))²
(m2 / m1) = (f1 / f2)²
Substituting the given values, we have:
(m2 / m1) = (14.2 / 3.90)²
Calculating the ratio:
(m2 / m1) = 56.84
Therefore, the ratio of masses, m2/m1, is approximately 56.84.
Please note that in this solution, we assumed that the spring constant does not change when the second object is added.