
A block is resting on a piston which executes simple harmonic motion with a period $2.0s$ . The maximum velocity of the piston, at an amplitude just sufficient for the block to separate from the piston, is
(A) $1.57m/s$
(B) $3.12m/s$
(C) $2.0m/s$
(D) $6.42m/s$
Answer
146.7k+ views
Hint It can be said that the maximum velocity of the piston is acquired when the block is just separated from the piston. And at this point, take the maximum acceleration of the block as the acceleration due to gravity and proceed.
Complete Step by step answer
Let $A$ denote the maximum amplitude of the piston. Let the maximum velocity of the piston be denoted by $v$ . Let the maximum acceleration be denoted by $a$ .
The block and the piston are initially together so we can say that both piston and block executes simple harmonic motion. As the block and the piston are not fixed to each other when the velocity of the piston is increased. At that instant, the acceleration of the block will be the acceleration due to gravity. So we can say that when the block acquires a maximum acceleration of $g$ (acceleration due to gravity), the block separates from the piston.
The maximum acceleration of the piston is given by
$a = A{\omega ^2}$
Or we can also write it as
$g = A{\omega ^2}$
The maximum velocity of the piston is given by
$v = A\omega $
By dividing the equation for maximum velocity and maximum acceleration, we can get a new equation for the maximum velocity of the piston as
$\dfrac{v}{g} = \dfrac{1}{\omega }$
$ \Rightarrow v = \dfrac{g}{\omega }$
It is given in the question that the time period of execution of simple harmonic motion by the piston is $2s$ . Therefore, we can calculate the angular frequency as
$\omega = \dfrac{{2\pi }}{T}$
$ \Rightarrow \omega = 3.14rad$
Substituting this value of angular frequency into the newly found equation for maximum velocity, we get
$v = \dfrac{{9.8}}{{3.14}} = 3.12m/s$
This is the maximum velocity acquired by the piston so that the block just separates from the piston.
Hence, option (B) is the correct option.
Note
An object executing simple harmonic motion will have its maximum amplitude as the maximum distance from the mean position and it will have the maximum velocity at the minimum distance from the mean position and will have the maximum acceleration at the maximum distance from the mean position.
Complete Step by step answer
Let $A$ denote the maximum amplitude of the piston. Let the maximum velocity of the piston be denoted by $v$ . Let the maximum acceleration be denoted by $a$ .
The block and the piston are initially together so we can say that both piston and block executes simple harmonic motion. As the block and the piston are not fixed to each other when the velocity of the piston is increased. At that instant, the acceleration of the block will be the acceleration due to gravity. So we can say that when the block acquires a maximum acceleration of $g$ (acceleration due to gravity), the block separates from the piston.
The maximum acceleration of the piston is given by
$a = A{\omega ^2}$
Or we can also write it as
$g = A{\omega ^2}$
The maximum velocity of the piston is given by
$v = A\omega $
By dividing the equation for maximum velocity and maximum acceleration, we can get a new equation for the maximum velocity of the piston as
$\dfrac{v}{g} = \dfrac{1}{\omega }$
$ \Rightarrow v = \dfrac{g}{\omega }$
It is given in the question that the time period of execution of simple harmonic motion by the piston is $2s$ . Therefore, we can calculate the angular frequency as
$\omega = \dfrac{{2\pi }}{T}$
$ \Rightarrow \omega = 3.14rad$
Substituting this value of angular frequency into the newly found equation for maximum velocity, we get
$v = \dfrac{{9.8}}{{3.14}} = 3.12m/s$
This is the maximum velocity acquired by the piston so that the block just separates from the piston.
Hence, option (B) is the correct option.
Note
An object executing simple harmonic motion will have its maximum amplitude as the maximum distance from the mean position and it will have the maximum velocity at the minimum distance from the mean position and will have the maximum acceleration at the maximum distance from the mean position.
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