Question

A mass m at the end of a spring vibrates with a frequency of 0.72 Hz . When an additional 700 g mass is added to m, the frequency is 0.64 Hz . Part A What is the value of m? Express your answer using two significant figures.

Answer 1

Answer:

The value of m is 2635.294 grams.

Explanation:

Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:

$f = \frac{\omega}{2\pi}$

Where $\omega$ is the angular frequency, measured in radians per second.

For a mass-spring system under simple harmonic motion, the angular frequency is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

The following equation is obtained after replacing angular frequency in frequency formula:

$f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }$

As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:

$f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}$

If $m_{2} = m_{1} + 700\,g$, $f_{1} = 0.72\,hz$ and $f_{2} = 0.64\,hz$, the resulting expression is simplified and then initial mass is found after clearing it:

$f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}$

$f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)$

$\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g$

$\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g$

$m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}$

$m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}$

$m_{1} = 2635.294\,g$

The value of m is 2635.294 grams.