Question

A block rides on a piston that is moving vertically with simple harmonic motion. The maximum speed of the piston is $2m/s$. At what amplitude of motion will the block and piston separate? ($g=10m/{{s}^{2}}$ )
$\begin{aligned} & \text{A}\text{.20cm} \\\ & \text{B}\text{.30cm} \\\ & \text{C}\text{.40cm} \\\ & \text{D}\text{.50cm} \\\ \end{aligned}$

Answer 1

Simple harmonic motion is a periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. To solve the question, we consider the two forces acting on the body and we equate them. And then the value of ‘g’, which is given in the question and the value of angular frequency, got by multiplying with amplitude and equating with velocity is substituted in the equation.

Formula used:
$\text{ }\\!\\!\omega\\!\\!\text{ A=v}$
Resting force=$\text{m}{{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}\text{A}$
Weight of the body=$\text{mg}$

Complete step by step answer:
In the question it is said that we have a block on a piston and the piston is moving vertically with simple harmonic motion.
The maximum speed of the piston is given, =$2m/s$.
We are asked to find the amplitude at which the piston and block will get separated.
Let us take ‘A’ as the amplitude.
The two forces acting here are resting force and weight of the body.
Resting force is given by the equation

Restoring force=$\text{m}{{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}\text{A}$

Where ‘m’ is the mass of the body$\text{ }\\!\\!\omega\\!\\!\text{ }$ is the angular frequency given by $\text{ }\\!\\!\omega\\!\\!\text{ =}\dfrac{\text{2 }\\!\\!\pi\\!\\!\text{ }}{\text{T}}$ and A is the amplitude.
Weight of the body is given by the equation
$\text{w=mg}$
Where m is the mass of the body and g is force of gravity given by$\text{10m/s}$.
When piston and block separate the resting force equals the weight of the body.
Hence we can equate the equations and

$\text{m}{{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}\text{A=mg}$
${{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}\text{A=g}$
$\text{A=}\dfrac{\text{g}}{{{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}}$

We know that amplitude multiplied by angular frequency gives velocity. Here velocity is$2m/s$. Therefore,
$\text{ }\\!\\!\omega\\!\\!\text{ A=2}$
Applying this in equation, we get

$\text{2 }\\!\\!\times\\!\\!\text{ }\\!\\!\omega\\!\\!\text{ =g}$
$\text{ }\\!\\!\omega\\!\\!\text{ =}\dfrac{\text{10}}{\text{2}}$
$\text{ }\\!\\!\omega\\!\\!\text{ =5rad/sec}$

Now to find the amplitude at which the piston and block separate, substitute the value of $\text{ }\\!\\!\omega\\!\\!\text{ }$and $\text{g}$ in equation

$\begin{aligned} & \text{A=}\dfrac{\text{g}}{{{\text{ }\\!\\!\omega\\!\\!\text{ }}^{\text{2}}}} \\\ & \text{A=}\dfrac{\text{10}}{{{\text{5}}^{\text{2}}}} \\\ & \text{A=}\dfrac{\text{10}}{\text{25}} \\\ & \text{A=0}\text{.4m} \\\ & \text{A=40cm} \\\ \end{aligned}$

Thus the amplitude is 40cm.
Hence, the correct answer is option C.

Note:
When the block riding on a piston gets separated, its resting force becomes equal to its weight. By this concept we can equate the two equations. To convert a quantity in meter to centimeter, multiply the quantity in meter by 100.