Question

Question: A block rides on a piston that is moving vertically with simple harmonic motion.

1. If the SHM has period 1.0 s, at what amplitude of motion will the block and piston separate?
2. If the piston has amplitude of 5.0 cm, what is the maximum frequency for which the block and piston will be in contact continuously?

Short answer

Answer:

1. The amplitude of motion at which the block and piston will separate if SHM has a time period 1.0 s is 0.25 m
2. The maximum frequency for which the block and piston will come in contact continuously if the piston has an amplitude of 5.0 cm is 2.2Hz

Step-by-step solution

The given data

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

Understanding the concept of simple harmonic motion

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)$\omega =\frac{2\pi }{T}$

The angular frequency of a body,$\omega =2\pi \text{f}$(ii)

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

a) Calculation of amplitude of motion

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

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

Piston and block separates when, a_m= g

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

$$\begin{gathered} x_m{\omega }^2=g......................................................\text{(a)} \\ {\left(\frac{2\pi }{T}\right)}^2{x}_m=g\text{(}\because \text{from equation (ii))} \\ {x}_m=\frac{g{T}^2}{4{\pi }^2} \\ =\frac{\left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right){\left(1.0\text{s}\right)}^2}{4{\pi }^2} \\ =0.25\text{m} \end{gathered}$$

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

b) Calculation of maximum frequency  

1. Using equations (a) & (ii) and the given value of amplitude of piston,

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

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.