A mass m at the end of a spring vibrates with a frequency of 0.93 Hz. When an additional 607 g mass is added to m, the frequency is 0.63 Hz. What is the value of m?
Frequency = [1/(2 pi)] sqrt (k/m)
0.93 = [1/(2 pi)] sqrt (k/m)
0.63 = [1/(2 pi)] sqrt [k/(m+0.607)]
Take the ratio. The k will cancel out.
0.93/0.63 = 1.476 = sqrt[(m+0.607)/m]
= sqrt (1 + 0.607/m )
2.179 = 1 + 0.607/m
1.170 = .607/m
Solve for m. It will be in kg since I wrote the added mass in kg
To determine the value of m, we need to use the formula for the frequency of a mass-spring system:
f = 1 / (2π) * √(k / m)
Where:
- f is the frequency of the system in Hz
- k is the spring constant
- m is the mass in kg
First, let's consider the initial situation, where only mass m is present. We can denote this as f₁ and m₁.
f₁ = 0.93 Hz (frequency)
m₁ = m (mass)
Next, let's consider the situation where an additional mass of 607 g (0.607 kg) is added to m. We can denote this as f₂ and m₂.
f₂ = 0.63 Hz (frequency)
m₂ = m + 0.607 kg (mass)
By comparing the two frequency equations, we can set up a ratio:
f₁ / f₂ = √(m₂ / m₁)
Substituting the given values:
0.93 / 0.63 = √((m + 0.607) / m)
Now, we can solve for m.
(0.93 / 0.63)² = (m + 0.607) / m
(0.93 / 0.63)² * m = m + 0.607
((0.93 / 0.63)² - 1) * m = 0.607
m = 0.607 / ((0.93 / 0.63)² - 1)
Calculating this expression provides the value of m.