91Ó°ÊÓ

One fly has a mass of m and the other fly has a mass of 2.5m.

The period T is the time required for one complete oscillation or cycle. It is related to the frequency by,

$\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

It is also related to mass (m) and force constant (k) by the formula,

$\mathit{T}\mathbf{=}\mathbf{2}\mathit{Ï€}\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

The angular frequencyis related to the period and frequency of the motion by,

$\mathit{Ó\neg }\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}}{\mathbf{T}}\mathbf{=}\mathbf{2}\mathit{Ï€}\mathit{f}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

Using the formula for the period of SHM, we can write the ratio of oscillation frequency for a fly with a mass ofm to a fly with a mass of 2.5m.

Formula:

The angular frequency of SHM, $\mathrm{Ó\neg }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ (i)

Since the spider is the same in both cases, the spring constant k of its web is the same. The fly has two different masses for the respective cases. Then, the ratio of oscillation frequency for the fly with mass m to mass 2.5m using equation (i) can be given as:

$$\begin{gathered} \frac{{\mathrm{Ó\neg }}_1}{{\mathrm{Ó\neg }}_2}=\frac{\sqrt{{\displaystyle \frac{k}{m_1}}}}{\sqrt{{\displaystyle \frac{k}{m_2}}}} \\ =\sqrt{\frac{m_2}{m_1}} \\ =\sqrt{\frac{2.5m}{m}} \\ =\sqrt{2.5} \\ =1.58 \end{gathered}$$

Therefore, the ratio of oscillation frequency for a fly with mass m to a fly with mass 2.5m is $1.58$.