When a 0.750-kg mass oscillates on an ideal spring, the frequency is 1.75 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Frequency of a SHM can be given as,

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\prod }_{}}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.................\left(1\right)$

Frequency of a SHM can be given as,

$k=4\prod _{}^2{f}^{2}m..................\left(2\right)$

For two different objects of mass $m_1$and $m_2$the frequencies will be different $f_{1}and{f}_2$respectively, but the spring constant being a constant will remain same

For mass $m_1$

$$\begin{gathered} k=4\prod _{}^2{f}^2{m}_1.................\left(3\right) \\ Formass{m}_2 \\ k=4\prod _{}^2{f}_2^2{m}_2.................\left(4\right) \end{gathered}$$

Equation (3) and (4) should be equal, so

$$\begin{gathered} k=4\prod _{}^2{f}_1^2{m}_1=4\prod _{}^2{f}_2^2{m}_2 \\ {f}_2=f\sqrt{\frac{m_1}{m_2}}...............\left(5\right) \end{gathered}$$

a)

$$\begin{gathered} m_1=0.75kg \\ {m}_2=0.75kg+0.220kg \\ =0.97kg \end{gathered}$$

For $m_1$frequency $f_1$= 1.75 Hz. Hence, from equation (5) we get

$$\begin{gathered} f_2=1.75Hz\times \sqrt{\frac{0.75kg}{0.97kg}} \\ {f}_2=1.54Hz \end{gathered}$$

b)

$$\begin{gathered} m_1=0.75kg \\ {m}_2=0.75kg-0.220kg \\ =0.53kg \end{gathered}$$

For frequency = 1.75 Hz. Hence, from equation (5) we get

$$\begin{gathered} f_2=1.75Hz\times \sqrt{\frac{0.75kg}{0.53kg}} \\ {f}_2=2.08Hz \end{gathered}$$

The frequency when 0.220 Kg added is 1.54 Hz and frequency when 0.220 Kg is subtracted is 2.08 Hz.