Answer
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Hint The SHM motion of the piston can be represented using a sinusoidal function. The speed of the piston can be calculated using the first time derivative of its position.
Complete step by step answer
We’ve been given that the piston in the cylinder of a locomotive is executing simple harmonic motion. Since it is executing simple harmonic motion, we can represent it using a sinusoidal function as follows:
$\Rightarrow x = A\sin (\omega t + \phi )$
Where $x$ is the position of the locomotive with respect to its origin, $\omega $ is the angular frequency, $t$ is the time and $\phi $ is the phase of the locomotive i.e. its relative position at the time $t = 0$ and $A$ is the amplitude of the piston
The velocity of the piston can then be calculated as:
$\Rightarrow v = \dfrac{{dx}}{{dt}}$
Substituting $x = A\sin (\omega t + \phi )$ in the above equation, we get
$\Rightarrow v = \dfrac{{d(A\sin (\omega t + \phi ))}}{{dt}}$
Taking the amplitude out of the derivative since it is a constant, we get
$\Rightarrow v = A\dfrac{d}{{dt}}\sin (\omega t + \phi ))$
$\Rightarrow v = Aw\cos (wt + \phi )$
Now that we know the velocity of the piston as a function of time, let's find out the maximum velocity. So, in the function of the velocity above, it will achieve the maximum velocity when the cosine term has the maximum value. We know that the cosine function has a maximum value of 1. So, the maximum velocity of the piston will be
$\Rightarrow v = A\omega $
Substituting the value of $A = 1/2$
Which gives, $A = 0.5\,m$
and $\omega = 200\,rad/\min $ in equation (1), we get
$\Rightarrow v = 0.5 \times 200$ which is the maximum speed of the piston
$\Rightarrow v = 100\,{\text{m}}/\min $.
Note
The maximum speed of the piston only depends on the amplitude of the piston and its angular frequency but the speed of the piston is actually a function of time. The piston will have the maximum speed when it passes through its equilibrium position as at that position it will have zero potential energy and maximum kinetic energy and hence maximum velocity.
Complete step by step answer
We’ve been given that the piston in the cylinder of a locomotive is executing simple harmonic motion. Since it is executing simple harmonic motion, we can represent it using a sinusoidal function as follows:
$\Rightarrow x = A\sin (\omega t + \phi )$
Where $x$ is the position of the locomotive with respect to its origin, $\omega $ is the angular frequency, $t$ is the time and $\phi $ is the phase of the locomotive i.e. its relative position at the time $t = 0$ and $A$ is the amplitude of the piston
The velocity of the piston can then be calculated as:
$\Rightarrow v = \dfrac{{dx}}{{dt}}$
Substituting $x = A\sin (\omega t + \phi )$ in the above equation, we get
$\Rightarrow v = \dfrac{{d(A\sin (\omega t + \phi ))}}{{dt}}$
Taking the amplitude out of the derivative since it is a constant, we get
$\Rightarrow v = A\dfrac{d}{{dt}}\sin (\omega t + \phi ))$
$\Rightarrow v = Aw\cos (wt + \phi )$
Now that we know the velocity of the piston as a function of time, let's find out the maximum velocity. So, in the function of the velocity above, it will achieve the maximum velocity when the cosine term has the maximum value. We know that the cosine function has a maximum value of 1. So, the maximum velocity of the piston will be
$\Rightarrow v = A\omega $
Substituting the value of $A = 1/2$
Which gives, $A = 0.5\,m$
and $\omega = 200\,rad/\min $ in equation (1), we get
$\Rightarrow v = 0.5 \times 200$ which is the maximum speed of the piston
$\Rightarrow v = 100\,{\text{m}}/\min $.
Note
The maximum speed of the piston only depends on the amplitude of the piston and its angular frequency but the speed of the piston is actually a function of time. The piston will have the maximum speed when it passes through its equilibrium position as at that position it will have zero potential energy and maximum kinetic energy and hence maximum velocity.
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