91Ó°ÊÓ

1. Period of motion, T = Is
2. Amplitude of piston,x_m= cm

The piston and block will separate when the acceleration is maximum, and it is equal to gravitational acceleration. Therefore, using the relation between angular velocity and the period of SHM, we can find the amplitude of motion at which the block and piston will separate if SHM has a period. Also, we can find maximum frequency using the relation between angular velocity and frequency.

Formulae:

The angular frequency of a body in motion,(i)$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

The angular frequency of a body,$\mathrm{Ó\neg }=2\mathrm{Ï€}\text{f}$(ii)

The acceleration of a body is directly proportional to displacement,${\text{a}}_m={\mathrm{Ó\neg }}^2{\text{x}}_m$(iii)

The general expression of position & acceleration of motion is given as:

$$\begin{gathered} x=x_m\cos \left(\mathrm{Ó\neg }t+f\right) \\ a=x_m{\mathrm{Ó\neg }}^2\cos \left(\mathrm{Ó\neg }t+f\right) \end{gathered}$$

Piston and block separates when, a_m= g

So using equation (iii), we get the acceleration as:

$$\begin{gathered} x_m{\mathrm{Ó\neg }}^2=g......................................................\text{(a)} \\ {\left(\frac{2\mathrm{Ï€}}{T}\right)}^2{x}_m=g\text{(}âˆ\mu \text{from equation (ii))} \\ {x}_m=\frac{g{T}^2}{4{\mathrm{Ï€}}^2} \\ =\frac{\left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right){\left(1.0\text{s}\right)}^2}{4{\mathrm{Ï€}}^2} \\ =0.25\text{m} \end{gathered}$$

Therefore, theamplitude of motion at which the block and piston will separate if the SHM has period 1.0 is 0.25.

$$\begin{gathered} x_m{\left(2\mathrm{Ï€}f\right)}^2=g \\ {x}_{m}4{\mathrm{Ï€}}^2{f}^2=g \\ {f}^2=\frac{g}{x_{m}4{\mathrm{Ï€}}^2} \\ f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{g}{x_m}} \\ =\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{9.8}{0.05\text{m}}} \\ =2.2\text{Hz} \end{gathered}$$

Therefore, maximum frequency for which the block and piston will be in contact continuously if the piston has the amplitude of 5.0 cm is 0.997 Hz.