91影视

Frequency of a SHM can be given as,

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\underset{}{\mathbf{鈭�}}}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.................\left(1\right)$

Frequency of a SHM can be given as,

$k=4\underset{}{\overset{2}{鈭�}}{f}^{2}m..................\left(2\right)$

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 $m_1$

$$\begin{gathered} k=4\underset{}{\overset{2}{鈭�}}{f}^2{m}_1.................\left(3\right) \\ Formass{m}_2 \\ k=4\underset{}{\overset{2}{鈭�}}{f}_2^2{m}_2.................\left(4\right) \end{gathered}$$

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

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

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

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

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

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

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