Question

an unknown mass, m, vibrates on a vertical spring with a frequency of 0.88 hz. when an additional 800 - g mass is added to m, the frequency is 0.48 hz.
what is the value of m?
what is the spring constant of the spring?

Explanation:

Step1: Write the formula for frequency

The formula for the frequency of a mass - spring system is \(f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\). So, \(f_1 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\) and \(f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m + 0.8}}\) (since \(800g=0.8kg\)).
From \(f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\), we can get \(k = 4\pi^{2}m f_{1}^{2}\). From \(f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m + 0.8}}\), we can get \(k=4\pi^{2}(m + 0.8)f_{2}^{2}\).

Step2: Equate the two expressions for \(k\)

Since \(4\pi^{2}m f_{1}^{2}=4\pi^{2}(m + 0.8)f_{2}^{2}\), we can cancel out \(4\pi^{2}\) on both sides. Then we have \(m f_{1}^{2}=(m + 0.8)f_{2}^{2}\).
Substitute \(f_1 = 0.88Hz\) and \(f_2=0.48Hz\) into the equation: \(m\times(0.88)^{2}=(m + 0.8)\times(0.48)^{2}\).
Expand the equation: \(0.7744m=0.2304m+0.18432\).
Subtract \(0.2304m\) from both sides: \(0.7744m-0.2304m=0.18432\).
\(0.544m = 0.18432\).
Solve for \(m\): \(m=\frac{0.18432}{0.544}=0.34kg\).

Step3: Calculate the spring constant \(k\)

Use \(k = 4\pi^{2}m f_{1}^{2}\). Substitute \(m = 0.34kg\) and \(f_1 = 0.88Hz\) into the formula.
\(k=4\pi^{2}\times0.34\times(0.88)^{2}\).
\(k = 4\times9.8696\times0.34\times0.7744\).
\(k\approx10.3N/m\).

Answer:

The value of \(m\) is \(0.34kg\) and the spring constant \(k\) is approximately \(10.3N/m\).